How Do You Know if a Integral Converges or Diverges

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Retrieve that the Fundamental Theorem of Calculus says that if \(f\) is a continuous function on the closed interval \([a,b]\text{,}\) then

\brainstorm{equation*} \ds{\int_a^b f(x)~dx=F(ten)\bigg|_a^b=F(b)-F(a)}\text{,} \end{equation*}

where \(F\) is any antiderivative of \(f\text{.}\)

Both the continuity condition and closed interval must hold to utilise the Primal Theorem of Calculus, and in this case, \(\ds\int_a^b f(x)\,dx\) represents the net expanse under \(f(x)\) from \(a\) to \(b\text{:}\)

We begin with an case where blindly applying the Fundamental Theorem of Calculus tin give an incorrect event.

Example 2.51. Using FTC.

Explain why \(\ds\int_{-ane}^1\frac{ane}{x^2}\,dx\) is not equal to \(-two\text{.}\)

Solution

Here is how ane might go along:

\brainstorm{equation*} \brainstorm{split} \int_{-one}^1\frac{1}{ten^2}\,dx \amp= \int_{-one}^1 ten^{-2}\,dx = -10^{-one}\bigg|_{-i}^1 = -\frac{1}{x}\bigg|_{-i}^1\\ \amp=~ \left(-\frac{1}{1}\correct) - \left(-\frac{1}{(-1)}\right) ~=~ -2\end{split} \end{equation*}

All the same, the above answer is WRONG! Since \(f(x)=i/x^2\) is non continuous on \([-1,~1]\text{,}\) nosotros cannot straight utilise the Fundamental Theorem of Calculus. Intuitively, we can run across why \(-two\) is not the right answer past looking at the graph of \(f(ten)=one/x^2\) on \([-1,~i]\text{.}\) The shaded surface area appears to grow without bound as seen in the effigy beneath.

Formalizing this case leads to the concept of an improper integral. There are 2 ways to extend the Fundamental Theorem of Calculus. One is to utilize an space interval, i.eastward., \([a,\infty)\text{,}\) \((-\infty,b]\) or \((-\infty,\infty)\text{.}\) The second is to let the interval \([a,b]\) to comprise an infinite discontinuity of \(f(x)\text{.}\) In either case, the integral is called an improper integral. One of the most of import applications of this concept is probability distributions because determining quantities similar the cumulative distribution or expected value typically require integrals on infinite intervals.

Subsection ii.7.1 Improper Integrals: Infinite Limits of Integration

To compute improper integrals, we use the concept of limits along with the Fundamental Theorem of Calculus.

Definition 2.52. Improper Integrals — One Infinite Limit of Integration.

If \(f(x)\) is continuous on \([a,\infty)\text{,}\) then the improper integral of \(f\) over \([a,\infty)\) is

\begin{equation*} \int_{a}^{\infty} f(x)\,dx=\lim_{R\to\infty}\int_a^R f(10)\,dx\text{.} \end{equation*}

If \(f(x)\) is continuous on \((-\infty,b]\text{,}\) then the improper integral of \(f\) over \((-\infty,b]\) is

\begin{equation*} \int_{-\infty}^b f(10)\,dx=\lim_{R\to -\infty}\int_R^b f(x)\,dx\text{.} \stop{equation*}

Since we are dealing with limits, we are interested in convergence and divergence of the improper integral. If the limit exists and is a finite number, nosotros say the improper integral converges. Otherwise, we say the improper integral diverges, which we capture in the following definition.

Definition 2.53. Convergence and Divergence.

If the limit exists and is a finite number, we say the improper integral converges.

If the limit is \(\pm\infty\) or does not exist, nosotros say the improper integral diverges.

To go an intuitive (though not completely correct) interpretation of improper integrals, nosotros endeavour to clarify \(\ds\int_a^\infty f(x)\,dx\) graphically. Here assume \(f(x)\) is continuous on \([a,\infty)\text{:}\)

Nosotros let \(R\) be a fixed number in \([a,\infty)\text{.}\) Then by taking the limit as \(R\) approaches \(\infty\text{,}\) nosotros get the improper integral:

\begin{equation*} \int_a^\infty f(x)\,dx=\lim_{R\to\infty}\int_a^R f(ten)\,dx\text{.} \terminate{equation*}

We tin can then use the Fundamental Theorem of Calculus to the last integral as \(f(10)\) is continuous on the closed interval \([a,R]\text{.}\)

We next ascertain the improper integral for the interval \((-\infty,~\infty)\text{.}\)

Definition two.54. Improper Integrals — Ii Infinite Limits of Integration.

If both \(\ds\int_{-\infty}^a f(10)\,dx\) and \(\ds\int_{a}^{\infty} f(x)\,dx\) are convergent, then the improper integral of \(f\) over \((-\infty,\infty)\) is

\begin{equation*} \int_{-\infty}^{\infty} f(ten)\,dx=\int_{-\infty}^a f(x)\,dx+\int_{a}^{\infty} f(x)\,dx \end{equation*}

The higher up definition requires both of the integrals

\begin{equation*} \int_{-\infty}^a f(ten)\,dx\qquad\mbox{and} \qquad\int_{a}^{\infty} f(10)\,dx \cease{equation*}

to be convergent for \(\ds\int_{-\infty}^{\infty} f(ten)\,dx\) to also be convergent. If either of \(\ds\int_{-\infty}^a f(x)\,dx\) or \(\ds\int_{a}^{\infty} f(10)\,dx\) is divergent, then then is \(\ds\int_{-\infty}^{\infty} f(x)\,dx\text{.}\)

Example 2.55. Improper Integral—I Infinite Limit of Integration.

Determine whether \(\ds\int_1^\infty\frac{ane}{10}\,dx\) is convergent or divergent.

Solution

Using the definition for improper integrals we write this as:

\begin{equation*} \begin{split} \int_1^\infty \frac{1}{10}\,dx\amp= \lim_{R\to\infty} \int_1^R\frac{one}{x}\,dx = \lim_{R\to\infty} \ln|x|\bigg|_1^R\\ \amp =\lim_{R\to\infty} \ln|R| - \ln|i| = \lim_{R\to\infty} \ln|R| = +\infty\end{separate} \stop{equation*}

Therefore, the integral is divergent.

Example 2.56. Improper Integral — Ii Infinite Limits of Integration.

Determine whether \(\ds\int_{-\infty}^\infty x\sin(x^2)\,dx\) is convergent or divergent.

Solution

We must compute both \(\ds\int_0^\infty x\sin(x^ii)\,dx\) and \(\ds\int_{-\infty}^0 x\sin(10^2)\,dx\text{.}\) Note that we don't have to divide the integral up at \(0\text{,}\) any finite value \(a\) will work. First nosotros compute the indefinite integral. Let \(u=ten^2\text{,}\) and so \(du=2x\,dx\) and hence,

\brainstorm{equation*} \int x\sin(x^two)\,dx=\frac{1}{2} \int \sin(u)\,du=-\frac{1}{2}\cos(x^ii)+C \end{equation*}

Using the definition of improper integral gives:

\begin{equation*} \brainstorm{split} \int_0^\infty x\sin(ten^two)\,dx \amp = \lim_{R\to\infty} \int_0^R ten\sin(x^2)\,dx =\lim_{R\to\infty} \left[-\frac{1}{2}\cos(ten^2)\right] \bigg|_0^R\\ \amp = -\frac{i}{two} \lim_{R\to\infty} \cos(R^two) +\frac{i}{2}\cease{split} \end{equation*}

This limit does not be since \(\cos x\) oscillates betwixt \(-1\) and \(+ane\text{.}\) In particular, \(\cos ten\) does not approach any particular value as \(x\) gets larger and larger. Thus, \(\ds\int_0^\infty x\sin(10^2)\,dx\) diverges, and hence, the integral \(\ds\int_{-\infty}^\infty ten\sin(x^2)\,dx\) diverges.

Subsection 2.7.2 Improper Integrals: Discontinuities

When there is a discontinuity in \([a,b]\) or at an endpoint, then the improper integral is as follows.

Definition ii.57.

{Improper Integrals — Discontinuities on Integration Bounds} If \(f(x)\) is continuous on \((a,b]\text{,}\) then the improper integral of \(f\) over \((a,b]\) is

\begin{equation*} \int_a^b f(ten)\,dx=\lim_{R\to a^+}\int_R^b f(x)\,dx\text{.} \end{equation*}

If \(f(x)\) is continuous on \([a,b)\text{,}\) then the improper integral of \(f\) over \([a,b)\) is

\begin{equation*} \int_a^b f(x)\,dx=\lim_{R\to b^-}\int_a^R f(x)\,dx\text{.} \end{equation*}

Definition 2.53 on convergence and divergence of an improper integral holds hither as well: If the limit above exists and is a finite number, nosotros say the improper integral converges. Otherwise, we say the improper integral diverges.

When there is a aperture in the interior of \([a,b]\text{,}\) we use the following definition.

Definition 2.58. Improper Integrals—Discontinuities Within Integration Interval.

If \(f\) has a discontinuity at \(x=c\) where \(c\in[a,b]\text{,}\) and both \(\ds\int_a^c f(ten)\,dx\) and \(\ds\int_c^b f(10)\,dx\) are convergent, so \(f\) over \([a,b]\) is

\begin{equation*} \int_a^b f(x)\,dx=\int_a^c f(10)\,dx+\int_c^b f(x)\,dx\text{.} \stop{equation*}

Again, nosotros can get an intuitive sense of this concept past analyzing \(\ds\int_a^b f(10)\,dx\) graphically. Here assume \(f(x)\) is continuous on \((a,b]\) but discontinuous at \(x=a\text{:}\)

We permit \(R\) exist a fixed number in \((a,b)\text{.}\) Then by taking the limit as \(R\) approaches \(a\) from the right, we get the improper integral:

\begin{equation*} \int_a^b f(x)\,dx=\lim_{R\to a^+}\int_R^b f(ten)\,dx\text{.} \terminate{equation*}

At present we tin utilise FTC to the terminal integral as \(f(x)\) is continuous on \([R,b]\text{.}\)

Example 2.59. A Divergent Integral.

Determine if \(\ds\int_{-one}^1\frac{1}{ten^2}\,dx\) is convergent or divergent.

Solution

The function \(f(x)=1/x^ii\) has a discontinuity at \(x=0\text{,}\) which lies in \([-1,ane]\text{.}\) We must compute \(\ds\int_{-1}^0 \frac{one}{x^2}\,dx\) and \(\ds\int_0^1 \frac{1}{x^2}\,dx\text{.}\) Permit's outset with \(\ds\int_0^1 \frac{ane}{ten^2}\,dx\text{:}\)

\brainstorm{equation*} \int_0^1 \frac{ane}{10^2}\,dx = \lim_{R\to 0^+} \int_R^one \frac{1}{x^2} \,dx = \lim_{R\to 0^+} -\frac{1}{x}\bigg|_R^1 = -1 + \lim_{R\to 0^+} \frac{1}{R} \end{equation*}

which diverges to \(+\infty\text{.}\) Therefore, \(\ds\int_{-one}^1\frac{1}{x^two}\,dx\) is divergent since i of \(\ds\int_{-1}^0 \frac{1}{x^ii}\,dx\) and \(\ds\int_0^1 \frac{1}{10^2}\,dx\) is divergent.

Example 2.sixty. Integral of the Logarithm.

Determine if \(\ds\int_0^ane \ln ten \,dx\) is convergent or divergent. Evaluate information technology if information technology is convergent.

Solution

Note that \(f(x)=\ln x\) is discontinuous at the endpoint \(x=0\text{.}\) We first use Integration by Parts to compute \(\ds\int\ln 10\,dx\text{.}\) We let \(u=\ln x\) and \(dv=dx\text{.}\) Then \(du=(1/ten)dx\text{,}\) \(v=ten\text{,}\) giving:

\begin{align*} \int \ln x\,dx \amp = \ds 10\ln x-\int x\cdot\frac{1}{x}\,dx\\ \amp = x\ln 10-\int 1\,dx\\ \amp = x\ln x-10+C \end{align*}

At present using the definition of improper integral for \(\ds\int_0^i \ln x \,dx\text{:}\)

\begin{equation*} \brainstorm{dissever} \int_0^i \ln x \,dx \amp= \lim_{R\to 0^+} \int_R^one \ln x\,dx = \lim_{R\to 0^+} (ten\ln x-x)\bigg|_R^1 \\ \amp= -i - \lim_{R\to 0^+}(R\ln R) + \lim_{R\to 0^+}R\terminate{carve up} \end{equation*}

Notation that \(\ds\lim_{R\to 0^+}R=0\text{.}\) We next compute \(\ds\lim_{R\to 0^+}(R\ln R)\text{.}\) Outset, we rewrite the expression equally follows:

\brainstorm{equation*} \lim_{ten\to0^+}(R\ln R)=\lim_{R\to0^+}\frac{\ln R}{i/R}\text{.} \end{equation*}

Now the limit is of the indeterminate type \((-\infty)/(\infty)\) and l'Hôpital's Rule can exist applied.

\begin{equation*} \lim_{R\to0^+}(R\ln R) =\lim_{R\to0^+}\frac{\ln R}{1/R} =\lim_{R\to0^+}\frac{1/R}{-i/R^two} =\lim_{R\to0^+}-\frac{R^2}{R} =\lim_{R\to0^+}(-R) =0 \stop{equation*}

Thus, \(\ds\lim_{R\to 0^+}(R\ln R)=0\text{.}\) Thus

\begin{equation*} \int_0^1 \ln ten \,dx = -ane\text{,} \end{equation*}

and the integral is convergent to \(-i\text{.}\)

Graphically, one might translate this to mean that the net area under \(\ln x\) on \([0,one]\) is \(-1\) (the area in this example lies beneath the \(x\)-axis).

Instance ii.61. Integral of a Square Root.

Determine if \(\ds\int_0^iv\frac{dx}{\sqrt{4-x}}\) is convergent or divergent. Evaluate information technology if it is convergent.

Solution

Notation that \(\frac{one}{\sqrt{4-x}}\) is discontinuous at the endpoint \(ten=iv\text{.}\) We use a \(u\)-substitution to compute \(\int \frac{dx}{\sqrt{4-ten}}\text{.}\) We let \(u=4-x\text{,}\) then \(du=-dx\text{,}\) giving:

\begin{align*} \ds\int\frac{dx}{\sqrt{four-ten}}\amp =\int-\frac{du}{u^{one/2}}\\ \amp =\int -u^{-one/2}\,du\\ \amp =-2(u)^{1/2}+C\\ \amp =-ii\sqrt{4-ten}+C \end{align*}

Now using the definition of improper integrals for \(\ds\int_0^4\frac{dx}{\sqrt{4-10}}\text{:}\)

\brainstorm{equation*} \ds\int_0^4\frac{dx}{\sqrt{4-x}}=\lim_{R\to4^{-}}(-2\sqrt{4-x})\bigg|_0^R=\lim_{R\to4^{-}}-2\sqrt{four-R}+2\sqrt{four}=4 \end{equation*}

Example 2.62. Improper Integral.

Decide if \(\ds\int_{one}^{two}\dfrac{dx}{\left( x-1\right) ^{1/three}}\) is convergent or divergent. Evaluate information technology if information technology is convergent.

Solution

Note that \(f\left( x\right) =\dfrac{i}{\left( ten-i\right) ^{1/3}}\) is discontinuous at the endpoint \(ten=1\text{.}\) We first utilise commutation to find \(\ds\int \dfrac{dx}{\left( ten-1\right) ^{1/three}}\text{.}\) We allow \(u=x-1\text{.}\) Then \(du=dx\text{,}\) giving

\begin{equation*} \int \dfrac{dx}{\left( ten-one\right) ^{ane/3}}=\int \frac{du}{u^{1/3}}=\int u^{-1/3}du=\frac{iii}{2}u^{2/3}+C=\frac{three}{2}\left( x-i\right) ^{2/3}+C\text{.} \terminate{equation*}

At present using the definition of improper integral for \(\ds\int_{ane}^{2}\dfrac{dx}{ \left( x-one\right) ^{ane/iii}}:\)

\begin{equation*} \begin{divide} \int_{i}^{2}\dfrac{dx}{\left( ten-ane\right) ^{1/iii}}\amp=\lim_{R\rightarrow 1^{+}}\int_{R}^{two}\dfrac{dx}{\left( ten-1\right) ^{one/3}}=\left. \lim_{R\rightarrow i^{+}}\frac{three}{2}\left( x-1\right) ^{2/3}\right\vert _{R}^{2} \\ \amp=\frac{3}{2}-\lim_{R\rightarrow 1^{+}}\frac{3}{2}\left( R-1\right) ^{2/three}=\frac{3}{2}\end{dissever}\text{,} \cease{equation*}

and the integral is convergent to \(\frac{3}{2}\text{.}\) Graphically, one might interpret this to hateful that the cyberspace area under \(\dfrac{i}{\left( 10-ane\right)^{1/iii}}\) on \(\left[ 1,2\right]\) is \(\frac{three}{two}\text{.}\)

Subsection 2.seven.three \(p\)-Integrals

Integrals of the course \(\ds \frac{1}{x^p}\) come up over again in the report of series. These integrals tin be either classified equally an improper integral with an infinite limit of integration, \(\ds\int_{a}^{\infty} \dfrac{1}{10^p}\,dx\text{,}\) or equally an improper integral with discontinuity at \(x=0\text{,}\) \(\ds\int_{0}^{a} \dfrac{i}{x^p}\,dx\text{.}\) In asymptotic analysis, it is useful to know when either of these intervals converge or diverge.

Theorem 2.63. \(p\)-Test for Infinite Limit.

For \(a>0\text{:}\)

1. If \(p>1\text{,}\) so \(\ds\int_{a}^{\infty }\frac{1}{x^p}\,dx\) converges.

2. If \(p\leq one\text{,}\) and so \(\ds\int_{a}^{\infty }\frac{1}{ten^p}~dx\) diverges.

Proof.

1. If \(p>ane\text{,}\) nosotros have \(\ds\int_{a}^{\infty }\frac{1}{10^p}\,dx=\lim_{R \rightarrow \infty} \left. \frac{ten^{1-p}}{i-p} \right|^R_a = \lim_{R\rightarrow \infty} \frac{R^{1-p}}{1-p} - \frac{a^{1-p}}{ane-p}=\frac{a^{ane-p}}{p-1}\text{.}\)

2. If \(p\leq ane\text{,}\) the to a higher place tells u.s.a. that the resulting limit is infinite.

Theorem 2.64. \(p\)-Test for Aperture.

For \(a>0\text{:}\)

1. If \(p\lt 1\text{,}\) then \(\ds\int_{0}^{a}\frac{one}{x^p}\,dx\) converges.

2. If \(p\geq 1\text{,}\) then \(\ds\int_{0}^{a}\frac{1}{10^p}\,dx\) diverges.

Proof.

1. If \(p\lt 1\text{,}\) we have to \(\ds\int_0^a \frac{1}{x^p}\, dx=\lim_{R \rightarrow 0^+} \left. \frac{x^{1-p}}{1-p} \right|^a_R = \lim_{R\rightarrow 0^+} \frac{a^{one-p}}{i-p} - \frac{R^{1-p}}{i-p}=\frac{a^{1-p}}{1-p}\text{.}\)

2. If \(p\geq 1\text{,}\) the to a higher place tells the states that the resulting limit is infinite.

With Case 2.55 and Example 2.59, you take already seen how the \(p\)-Test is applied. For proficient measure, here is one more than example.

Case ii.65. \(p\)-Test.

Determine if the following integrals are convergent or divergent.

1. \(\ds\int_1^{\infty} \frac{1}{10^3}\,dx\)

2. \(\ds\int_0^{5} \frac{1}{x^4}\,dx\)

Solution

1. This is a \(p\)-integral with an infinite upper limit of integration and \(p=3 > 1\text{.}\) Therefore, by the \(p\)-Test for Space Limit, \(\ds\int_1^{\infty} \frac{1}{10^3}\,dx\) converges.

2. We classify \(\ds\int_0^{5} \frac{1}{ten^4}\,dx\) as a \(p\)-integral with a discontinuity at \(x=0\) and \(p=4 \geq i\text{.}\) Thus, past the \(p\)-Test for Discontinuity, the integral diverges.

Subsection 2.vii.4 Comparison Test

The following test allows us to determine convergence/difference information nigh improper integrals that are hard to compute by comparing them to easier ones. Nosotros state the test for \([a,\infty)\text{,}\) but like versions hold for the other improper integrals.

Theorem two.66. Comparison Exam for Improper Integrals.

Presume that \(f(ten)\geq yard(x)\geq 0\) for \(x\geq a\text{.}\)

1. If \(\ds\int_a^\infty f(ten)\,dx\) \thmfont{converges}, and then \(\ds\int_a^\infty g(x)\,dx\) also \thmfont{converges}.

2. If \(\ds\int_a^\infty g(x)\,dx\) \thmfont{diverges}, then \(\ds\int_a^\infty f(10)\,dx\) also \thmfont{diverges}.

Informally, (i) says that if \(f(x)\) is larger than \(g(ten)\text{,}\) and the area nether \(f(x)\) is finite (converges), then the surface area under \(g(10)\) must likewise exist finite (converges). Informally, (ii) says that if \(f(x)\) is larger than \(g(ten)\text{,}\) and the area nether \(g(x)\) is infinite (diverges), then the area under \(f(ten)\) must also be space (diverges).

Example 2.67. Comparing Test.

Bear witness that \(\ds\int_2^\infty \frac{\cos^2x}{ten^2} \,dx\) converges.

Solution

We utilise the Comparison Test to show that information technology converges. Notation that \(0\leq \cos^2x\leq 1\) and hence

\begin{equation*} 0 \leq\frac{\cos^2x}{x^2}\leq\frac{one}{ten^2}\text{.} \end{equation*}

Thus, taking \(f(10)=1/x^2\) and \(g(ten)=\cos^2x / 10^2\) we have \(f(10)\geq thou(x)\geq 0\text{.}\) 1 can easily see that \(\ds\int_2^\infty \frac{1}{ten^ii}\,dx\) converges. Therefore, \(\ds\int_2^\infty \frac{\cos^2x}{ten^ii} \,dx\) also converges.

Exercises for Section two.vii.

Exercise 2.7.ane.

Determine whether the following improper integrals are convergent or divergent. Evaluate those that are convergent.

1. \(\ds\int_{0}^{\infty}\dfrac{one}{10^2+1}\,dx\)

   Respond Solution

   We first compute the corresponding indefinite integral using trigonometric substitution. Allow \(x = \tan \theta\) and so \(dx = \sec^2\theta\,d\theta\text{:}\)

   \brainstorm{equation*} \begin{split} \int \frac{ane}{x^2+1}\,dx \amp = \int\frac{1}{1+\tan^2\theta} \sec^2\theta\,d\theta\\ \amp = \int \frac{\sec^2\theta}{\sec^ii\theta}\,d\theta\\ \amp = \int\,d\theta\\ \amp = \theta + C\\ \amp = \arctan(x)+ C \end{split} \end{equation*}

   We recall the graph of \(y=\arctan(x)\text{:}\)

   

   Therefore,

   \begin{equation*} \begin{split} \int_0^{\infty} \frac{1}{x^2+i}\,dx \amp = \lim_{R\to\infty} \int_0^R \frac{1}{x^ii+1}\,dx \\ \amp = \lim_{R\to\infty} \arctan(R) - \arctan(0)\\ \amp =\frac{\pi}{ii} - 0\\ \amp = \frac{\pi}{2}. \end{split up} \finish{equation*}

   Hence the interval is converges to \(\pi/2\text{.}\)

2. \(\ds\int_{0}^{\infty}\dfrac{x}{10^2+ane}\,dx\)

   Answer

   Divergent (to \(\infty\))

   Solution

   We kickoff solve the respective indefinite integral using the following substitution:

   \begin{equation*} u = 1+x^2, du = 2x\,dx\text{.} \end{equation*}

   And then

   \begin{equation*} \begin{split} \int \frac{x}{10^2+1}\,dx \amp = \int \frac{1}{u} \frac{du}{ii}\\ \amp = \frac{1}{2} \int\frac{ane}{u}\,du \\ \amp = \frac{one}{2} \ln |u| + C\\ \amp = \frac{ane}{2} \ln |x^ii+1| + C \end{dissever} \end{equation*}

   Therefore,

   \begin{equation*} \begin{carve up} \int_0^\infty \frac{10}{1+x^ii}\,dx \amp = \lim_{R\to\infty}\int_0^R \frac{10}{1+x^2}\,dx \\ \amp = \frac{1}{2}\left[\lim_{R\to \infty} \ln|R^2 + 1| - \ln(1)\right] \end{carve up} \stop{equation*}

   Hence, the integral diverges.

3. \(\ds\int_{0}^{\infty}eastward^{-10}(\cos ten+\sin x)\,dx\text{.}\)

   Respond Solution

   We first find the corresponding indefinite integral using Integration by Parts: Let

   \begin{equation*} \begin{array}{cc} u = e^{-10} \amp dv = (\cos x + \sin x)\,dx \\ du = -e^{-x} \, dx \amp v = \sin x - \cos x. \end{array} \end{equation*}

   Therefore,

   \begin{equation*} \int e^{-x}\left(\cos x + \sin 10\right)\, dx = due east^{-x}(\sin x - \cos x) + \int due east^{-ten}\left(\sin x - \cos x\right)\,dx\text{.} \end{equation*}

   Now allow

   \begin{equation*} \begin{assortment}{cc} u = e^{-10} \amp dv = (\sin x - \cos x)\,dx \\ du = -due east^{-10} \, dx \amp five = -\left(\sin ten + \cos x\right), \end{array} \end{equation*}

   which gives

   \begin{equation*} \begin{divide} \int eastward^{-x}\left(\cos x + \sin x\right)\, dx \amp = -two e^{-10}\cos ten - \int due east^{-ten}\left(\cos x + \sin x\right)\, dx \\ 2\int east^{-ten}\left(\cos x + \sin x\right)\, dx \amp = -ii e^{-ten}\cos x \\ \int e^{-x}\left(\cos x + \sin 10\right)\, dx \amp = -e^{-10}\cos x + C. \terminate{split} \end{equation*}

   Therefore,

   \brainstorm{equation*} \brainstorm{split} \int_0^\infty e^{-ten}\left(\cos x + \sin x\right) \, dx \amp = \lim_{R\to\infty^-} \int_0^R e^{-x}\left(\cos x + \sin x\right) \, dx\\ \amp =\lim_{R \to \infty^{-}} -e^{-x}\cos 10 \big\vert_0^{R} \\ \amp = \lim_{R \to \infty^{-}} -due east^{-R}\cos R - (-1)\\ \amp = i, \end{separate} \cease{equation*}

   i.e. the integral converges to 1.

4. \(\ds\int_{0}^{\pi/ii}\sec^{2}x\,dx\)

   Answer

   Divergent (to \(\infty\))

   Solution

   The integrand \(f(x) = \sec^2(ten)\) has vertical asymptotes at \(x=ii\pi chiliad \pm \frac{\pi}{ii}\text{,}\) and so is divers on \​(\left[0, \frac{\pi}{2}\right)\), just has a discontinuity at the endpoint \​(x=\frac{\pi}{2}\). Nosotros recall that \​(\diff{}{x} \tan(10) = \sec^2 (10)\). Hence,

   \brainstorm{equation*} \begin{split} \int_0^{\pi/2} \sec^2(x) \, dx \amp = \lim_{R \to \frac{\pi}{2}^-} \int_0^R \sec^2(ten)\,dx \\ \amp = \lim_{R \to \frac{\pi}{2}^-} \tan(x)\bigg\lvert_0^R\\ \amp = \lim_{R \to \frac{\pi}{2}^-} \tan(R). \end{carve up} \cease{equation*}

   

   From the graph of \(y=\tan(ten)\) in a higher place, we conclude that

   \brainstorm{equation*} \lim_{R \to \frac{\pi}{ii}^-} \tan(R) = +\infty\text{,} \end{equation*}

   and and then the integral diverges.

5. \(\ds\int_{0}^{4}\dfrac{1}{(iv-10)^{2/5}}\,dx\)

   Reply

   Converges to \(\frac{5}{3}(4^{3/five})\)

   Solution

   The integrand \(f(ten) = \frac{ane}{(4-x)^{2/5}}\) has a discontinuity at the correct endpoint \(x=4\text{.}\) Therefore,

   \begin{equation*} \brainstorm{dissever} \int_0^iv \frac{i}{(4-ten)^{2/5}} \, dx \amp = \lim_{R\to iv^-} \int_0^R \frac{ane}{(4-x)^{2/v}}\,dx \\ \amp = \lim_{R\to 4^-} \left(-\frac{five}{3}(4-x)^{3/5}\correct) \bigg\lvert_0^R\\ \amp = \left(\frac{5}{3} \lim_{R\to iv^-} (4-R)^{iii/5}\correct) - \left( -\frac{5}{iii} 4^{3/five} \right)\\ \amp = 0 + \frac{5}{3} 4^{3/v}. \end{dissever} \end{equation*}

   Therefore, the integral converges to \(\frac{5}{three} iv^{3/v}\text{.}\)

6. \(\ds\int_1^{\infty}\frac{1}{10^2}\,dx\)

   Reply Solution

   This is a \(p\)-integral with an infinite upper limit and \(p=iii > 1\text{.}\) Therefore, past the \(p\)-exam for Infinte Limit, the integral converges.

   \begin{equation*} \brainstorm{split} \int_1^{\infty} \frac{i}{x^two} \, dx \amp = \lim_{R\to\infty} \int_1^R \frac{1}{ten^2}\,dx \\ \amp = \lim_{R\to\infty} \frac{-1}{x}\bigg\lvert_1^R \\ \amp = \left(- \lim_{R\to\infty} \frac{i}{R}\right) - \left(-1\right)\\ \amp = 0 + 1 \end{split} \end{equation*}

   Therefore, the integral converges to one.

7. \(\ds\int_e^{\infty}\frac{one}{x\sqrt{\ln 10}}\,dx\)

   Answer Solution

   The integrand \(f(x) = \frac{1}{x\sqrt{\ln x}}\) is continuous on \([e,\infty)\text{.}\) We first compute the corresponding indefinite integral using a commutation: Permit \(u=\ln x\text{,}\) with \(du = \frac{1}{x}\,dx\text{.}\)

   \begin{equation*} \begin{split} \int \frac{1}{10\sqrt{\ln x}} \, dx \amp = \int \frac{one}{\sqrt{u}}\,du\\ \amp = 2\sqrt{u} + C\\ \amp = 2\sqrt{\ln(10)} + C \end{split} \stop{equation*}

   Therefore, we have

   \begin{equation*} \begin{dissever} \int_e^{\infty} \frac{i}{10\sqrt{\ln x}} \, dx \amp = \lim_{R\to\infty} \int_e^R \frac{1}{x\sqrt{\ln x}} \, dx \\ \amp = \lim_{R\to\infty} two\sqrt{\ln(ten)} \bigg\lvert_e^R\\ \amp = 2\left(\lim_{R\to\infty} \sqrt{\ln(R)} - 1\right)\\ \amp = \infty \end{split} \end{equation*}

   Hence, the integral diverges.

8. \(\ds\int_0^{\infty}e^{-3x}\,dx\)

   Reply Solution

   Nosotros compute:

   \begin{equation*} \begin{divide} \int_0^{\infty} e^{-3x} \, dx \amp = \lim_{R\to\infty} \int_0^R e^{-3x}\,dx \\ \amp = -\frac{1}{iii} \lim_{R\to\infty} e^{-3x} \bigg\lvert_0^R\\ \amp = -\frac{1}{iii} \left(0 - i\right)\\ \amp = \frac{1}{iii}. \cease{carve up} \end{equation*}

   Therefore, the integral converges to \(1/three\text{.}\)

9. \(\ds\int_1^east\frac{1}{x(\ln x)^ii}\,dx\)

   Answer Solution

   The integrand \(f(x) = \frac{one}{x(\ln x)^2}\) has a discontinuity at \(x=1\text{.}\) We first compute the corresponding indefinite integral. Let \(u=\ln x\) with \(du = \frac{1}{x}\,dx\text{:}\)

   \begin{equation*} \begin{split} \int \frac{1}{ten(\ln x)^ii}\,dx \amp = \int \frac{1}{u^2}\,du\\ \amp = -\frac{1}{u} + C\\ \amp = -\frac{1}{\ln x} + C \end{split} \end{equation*}

   We now compute the definite integral:

   \begin{equation*} \brainstorm{separate} \int_1^e \frac{1}{10(\ln 10)^2}\,dx \amp = \lim_{R\to 1^+} \int_R^east\frac{one}{ten(\ln x)^2}\,dx \\ \amp = \lim_{R\to 1^+} \left[-\frac{ane}{\ln x} \right]_R^e\\ \amp = -ane - \left(- \lim_{R\to 1^+} \frac{ane}{\ln R}\right) \end{split} \end{equation*}

   We conclude that the integral diverges (to \(+\infty\)).

10. \(\ds\int_0^{\infty}e^{-x}\sin^ii\left(\frac{\pi x}{2}\right)\,dx\)

    Respond

    Converges to \(\frac{\pi^2}{two+ii\pi^2}\)

    Solution

    First, nosotros rewrite

    \brainstorm{equation*} \sin^2\left(\frac{\pi 10}{2}\right) = \frac{1}{2}\left(one-\cos(\pi 10)\right)\text{.} \finish{equation*}

    Therefore,

    \begin{equation*} \int_0^{\infty} e^{-x}\sin^two\left(\frac{\pi x}{two}\right) \, dx = \frac{i}{2}\left(\int_0^{\infty} e^{-x}\,dx -\int_0^{\infty} e^{-x}\cos(\pi x) \, dx\right) \end{equation*}

    We will commencement integral \(\displaystyle \int due east^{-x}\cos(\pi ten)\,dx\) using Integration by Parts: Let

    \begin{equation*} \begin{array}{cc} u = e^{-ten} \amp dv = \cos(\pi x)\,dx \\ du = -e^{-10} \, dx \amp v = \sin(\pi x)/\pi. \terminate{array} \end{equation*}

    This gives

    \brainstorm{equation*} \int e^{-x}\cos(\pi x)\,dx = \frac{east^{-x}\sin(\pi ten)}{\pi} + \frac{one}{\pi} \int e^{-ten}\sin(\pi ten)\,dx\text{.} \terminate{equation*}

    We at present integrate by parts one more than time, with:

    \begin{equation*} \brainstorm{array}{cc} u = e^{-ten} \amp dv = \sin(\pi x)\,dx \\ du = -e^{-ten} \, dx \amp v = -\cos(\pi ten)/\pi. \end{array} \end{equation*}

    Thus,

    \begin{equation*} \int e^{-10}\cos(\pi x)\,dx = \frac{eastward^{-x}\sin(\pi x)}{\pi} - \frac{e^{-ten}\cos(\pi x)}{\pi ^ii} - \frac{1}{\pi^2} \int e^{-x}\cos(\pi ten)\,dx\text{.} \end{equation*}

    We now notice that the integral appears on both sides of the equation, and and so we can combine terms:

    \brainstorm{equation*} \begin{split} \pi ^ii \int e^{-10}\cos(\pi x)\,dx \amp = \pi due east^{-x}\sin(\pi x) - e^{-x}\cos(\pi x) - \int east^{-x}\cos(\pi 10)\,dx\\ \left(\pi^2 + 1\right) \int eastward^{-ten}\cos(\pi x)\,dx \amp = \pi eastward^{-10}\sin(\pi ten) - e^{-x}\cos(\pi x)\\ \int e^{-x}\cos(\pi x)\,dx \amp = \frac{\pi e^{-x}\sin(\pi 10) - eastward^{-10}\cos(\pi x)}{\pi^2 + one}. \end{split} \cease{equation*}

    All together, we observe that

    \begin{equation*} \begin{carve up} \int eastward^{-x}\sin^2\left(\frac{\pi ten}{ii}\right) \, dx \amp = \frac{1}{2}\left(\int due east^{-x}\,dx -\int e^{-x}\cos(\pi ten) \, dx\right)\\ \amp = -\frac{e^{-x}}{ii} - \frac{i}{two}\left[\frac{\pi e^{-x}\sin(\pi x) - e^{-x}\cos(\pi 10)} {\pi^two + 1} \right] \end{split} \end{equation*}

    We at present solve the definite integral:

    \brainstorm{equation*} \brainstorm{split} \int_0^{\infty} e^{-ten}\sin^2\left(\frac{\pi ten}{2}\correct) \,dx \amp = \lim_{R\to\infty} \int_0^R due east^{-10}\sin^2\left(\frac{\pi x}{two}\right) \,dx\\ \amp = \lim_{R\to\infty} \left[ -\frac{e^{-10}}{2} - \frac{i}{ii}\left(\frac{\pi east^{-ten}\sin(\pi 10) - e^{-x}\cos(\pi x)} {\pi^two + one} \correct)\right]_0^R\\ \amp = 0 - \left[-\frac{1}{2} - \frac{1}{ii}\left(\frac{-i}{\pi^two+1}\correct) \correct]\\ \amp = \frac{1}{2} - \frac{1}{\pi^ii+1}. \stop{divide} \end{equation*}

    Therefore, the integral converges to \(\frac{1}{2} - \frac{1}{\pi^2+ane}\text{.}\)

11. \(\ds\int_{-\infty}^{\infty}\frac{one}{ten^2+ane}\,dx\)

    Answer Solution

    We first compute the corresponding indefinite integral using trigonometric substitution. Let \(x = \tan \theta\) and and then \(dx = \sec^2\theta\,d\theta\text{:}\)

    \begin{equation*} \begin{carve up} \int \frac{1}{x^2+1}\,dx \amp = \int\frac{ane}{1+\tan^two\theta} \sec^ii\theta\,d\theta\\ \amp = \int \frac{\sec^ii\theta}{\sec^2\theta}\,d\theta\\ \amp = \int\,d\theta\\ \amp = \theta + C\\ \amp = \arctan(10)+ C \terminate{split up} \end{equation*}

    We recall the graph of \(y=\arctan(x)\text{:}\)

    

    Using the fact that the integrand is an fifty-fifty role, we calculate

    \begin{equation*} \begin{divide} \int_{-\infty}^{\infty} \frac{1}{10^2+1}\,dx \amp = 2 \int_0^{\infty} \frac{1}{x^two+ane}\,dx \\[1ex] \amp = two \lim_{R\to\infty^-} \int_0^R \frac{1}{ten^2+1}\,dx \\[1ex] \amp = 2 \lim_{R\to\infty^-} \arctan x \big\vert_0^R \\[1ex] \amp =2 \lim_{R\to\infty^-} \arctan R - \arctan 0 \\[1ex] \amp = 2 \left(\frac{\pi}{2} -0\correct) = \pi. \end{split} \end{equation*}

12. \(\ds\int_{-\infty}^{\infty}\frac{10}{x^2+1}\,dx\)

    Reply Solution

    Nosotros first find the corresponding indefinite integral past making the exchange,

    \begin{equation*} u = 10^2 +ane,\ du = 2x\,dx\text{.} \end{equation*}

    \begin{equation*} \int\frac{x}{x^2+1}\,dx= \int \frac{ane}{u} \frac{du}{2} = \frac{1}{2} \ln|x^2 + one| + C\text{.} \cease{equation*}

    We kickoff consider the integral

    \begin{equation*} \begin{split up} \int_0^\infty \frac{10}{x^2+ane}\,dx \amp = \lim_{R\to\infty^-} \int_0^R \frac{x}{x^2+1}\,dx \\ \amp = \lim_{R\to\infty^-} \frac{1}{2} \ln|ten^two + one| \big\vert_0^R \\ \amp = \frac{i}{2} \lim_{R\to\infty^-} \ln|R^2+one| - \frac{1}{2} \ln |1| \to \infty. \end{split up} \end{equation*}

    And then since this integral is divergent, we must have that \(\displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2+1}\,dx\) is divergent.

Exercise 2.vii.2.

Prove that the integral \(\ds{\int_{1}^{\infty}\frac{1}{x^p}\,dx}\) is convergent if \(p>1\) and divergent if \(0\lt p\leq 1\text{.}\)

Solution

Since

\begin{equation*} \int_1^{\infty}\frac{one}{10^p}\,dx = \lim_{R\to\infty} \frac{x^{1-p}}{i-p} \bigg\lvert_1^R = \lim_{R\to\infty} \frac{R^{1-p}}{1-p} - \frac{1}{ane-p}\text{.} \cease{equation*}

So if \(p>ane\text{,}\) nosotros see that the integral converges. However for \(0 \lt p \leq 1\text{,}\) the integral diverges.

Exercise 2.7.three.

Suppose that \(p>0\text{.}\) Find all values of \(p\) for which \(\ds{\int_{0}^{ane}\dfrac{1}{10^p}\,dx}\) converges.

Respond Solution

We wish to find all values of \(p\) for which the integral \(\displaystyle \int_0^1 \frac{one}{ten^p} \,dx\) converges. We know that

\begin{equation*} \int_0^i \frac{1}{10}\,dx = \lim_{R\to 0^+} \int_R^i \frac{1}{x}\,dx = \lim_{R\to 0^+} \ln x \big\vert_R^1 = \ln (i) - \lim_{R\to 0^+} \ln R \to \infty\text{,} \end{equation*}

and and then for \(p=1\) the integral diverges. For \(p > 0, \neq i\text{,}\) nosotros have

\brainstorm{equation*} \int \frac{i}{ten^p} \,dx = \frac{x^{1-p}}{ane-p} + C\text{.} \stop{equation*}

We are left with two cases: If \(p > 1\text{,}\) so

\begin{equation*} \int_0^1 \frac{i}{ten^p}\,dx = \lim_{R\to 0^+} \int_R^i \frac{i}{x}\,dx =\lim_{R\to 0^+} \frac{x^{1-p}}{1-p} \bigg\vert_R^ane = \frac{-i}{p-1}+ \frac{one}{p-1}\lim_{R\to 0^+} \frac{1}{R^{p-i}}, \qquad (p-one > 0) \end{equation*}

which diverges. If \(\lt p \lt 1\text{,}\) and then

\begin{equation*} \int_0^1 \frac{1}{10^p}\,dx = \lim_{R\to 0^+} \int_R^1 \frac{1}{x}\,dx = \lim_{R\to 0^+} \frac{ten^{1-p}}{1-p} \bigg\vert_R^1 = \frac{one}{i-p}- \frac{1}{1-p}\lim_{R\to 0^+} R^{ane-p}, \qquad (i-p > 0) \cease{equation*}

which converges to \(\dfrac{1}{1-p}\text{.}\) Thus, the integral converges for all \(p \in (0,1)\text{,}\) and diverges for all \(p\geq 1\text{.}\)

Exercise 2.seven.iv.

Show that \(\ds{\int_{1}^{\infty}\dfrac{\sin^2 ten}{x(\sqrt{x}+ane)}\,dx}\) converges.

Solution

Nosotros now bear witness that \(\displaystyle \int_1^\infty \frac{\sin^2 x}{x(\sqrt{x}+1)} \,dx\) converges using the comparison test. First notice that \(0 \leq \sin^2 x \leq one\) for all \(10\text{.}\) Therefore,

\begin{equation*} \frac{\sin^two x}{x(\sqrt{ten}+1)} \leq \frac{1}{ten(\sqrt{x}+1)} = \frac{ane}{ten^{three/ii}+x} \leq \frac{ane}{x^{iii/ii}}, x \in[1,\infty)\text{.} \cease{equation*}

Since

\begin{equation*} \int_1^\infty \frac{one}{x^{3/2}}\,dx = \lim_{R\to\infty^-} \int_1^R \frac{1}{x^{iii/2}}\,dx = \lim_{R\to\infty^-}\frac{-2}{\sqrt{x}} \bigg\vert_1^R = \lim_{R\to\infty^-} \frac{-2}{\sqrt R} - (-2) = 2\text{,} \finish{equation*}

we conclude that \(\displaystyle \int_1^\infty \frac{\sin^two ten}{x(\sqrt{x}+one)} \,dx\) converges.

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Source: https://www.sfu.ca/math-coursenotes/Math%20158%20Course%20Notes/sec_ImproperIntegrals.html

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