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\int_{0}^{+\infty }\frac{1}{x^{6}+1}dx Compute

Question:

Compute

{eq}\int_{0}^{+\infty }\frac{1}{x^{6}+1}dx {/eq}

Definite Integral

Given an integral of the form {eq}I=\int_{x=-\infty}^\infty f(x)~dx {/eq} where {eq}f(x) {/eq} is a real valued function, it is possible to use a version of the residue formula from complex analysis to evaluate the integral. One form of the residue formula states the following.

Residue Formula: Let {eq}f(z) {/eq} be analytic in {eq}\mathbb{C} {/eq} except for a finite number of poles, none of which are not on the real axis. If {eq}f(z)=\frac{p(z)}{q(z)} {/eq} where {eq}p(z) {/eq} and {eq}q(z) {/eq} are polynomials with {eq}\deg(q(z))\geq2+\deg(p(z)) {/eq} then

{eq}\begin{eqnarray*} I &=& \int_{x=-\infty}^\infty f(x)~dx \\ &=& 2\pi i\sum_{k}\text{res}(f,z_k) \end{eqnarray*} {/eq}

where {eq}z_k {/eq} is a pole of {eq}f {/eq} in the upper half of the complex plane and {eq}\text{res}(f,z_k) {/eq} is the residue of {eq}f {/eq} at {eq}z_k {/eq}. There are several formulas for calculating the residues. One such formula will be used in the answer that follows.

Answer and Explanation:

The question is restated with slightly different notation. Compute

{eq}\begin{eqnarray*} I_1 = \int_{x=0}^{\infty}\frac{1}{x^{6}+1}~dx. \end{eqnarray*} {/eq}

The following form of the residue formula from complex analysis will be used to evaluate the integral.

Residue Formula: Let {eq}f(z) {/eq} be analytic in {eq}\mathbb{C} {/eq} except for a finite number of poles, none of which are not on the real axis. If {eq}f(z)=\frac{p(z)}{q(z)} {/eq} where {eq}p(z) {/eq} and {eq}q(z) {/eq} are polynomials with {eq}\deg(q(z))\geq2+\deg(p(z)) {/eq} then

{eq}\begin{eqnarray*} I &=& \int_{x=-\infty}^\infty f(x)~dx \\ &=& 2\pi i\sum_{k}\text{res}(f,z_k) \end{eqnarray*} {/eq}

where {eq}z_k {/eq} is a pole of {eq}f {/eq} in the upper half of the complex plane and {eq}\text{res}(f,z_k) {/eq} is the residue of {eq}f {/eq} at {eq}z_k {/eq}.

To use the residue formula we consider the complex valued function {eq}f(z)=\frac{1}{z^6+1} {/eq}. Note that

{eq}\begin{eqnarray*} f(z) &=& \frac{p(z)}{q(z)} \\ p(z) &=& 1 \\ q(z) &=& z^6+1 \\ \deg(p(z)) &=& 0 \\ \deg(q(z)) &=& 6. \end{eqnarray*} {/eq}

Consequently, {eq}\deg(q(z))\geq2+\deg(p(z)) {/eq}. The sixth roots of {eq}q(z)=z^6+1 {/eq} are {eq}z_k=e^\frac{(2k+1)\pi i}{6} {/eq} for {eq}k=0,1,2,3,4,5 {/eq}. These are all simple poles of {eq}f(z) {/eq}. Of these poles, {eq}z_k=e^\frac{(2k+1)\pi i}{6} {/eq} for {eq}k=0,1,2 {/eq} are in the upper half of the complex plane. The other three poles are in the lower half of the complex plane. Therefore, the conditions of the residue formula hold. By the residue formula we get

{eq}\begin{eqnarray*} I &=& \int_{x=-\infty}^\infty \frac{1}{x^6+1}~dx \\ &=& 2\pi i\sum_{k=0}^2\text{res}(f,z_k) \\ &=& 2\pi i~(\text{res}(f,z_0)+\text{res}(f,z_1)+\text{res}(f,z_2)) \\ &=& 2\pi i\left(\text{res}\left(f,e^\frac{\pi i}{6}\right)+\text{res}\left(f,e^\frac{3\pi i}{6}\right) +\text{res}\left(f,e^\frac{5\pi i}{6}\right)\right) \\ &=& 2\pi i\left(\text{res}\left(f,e^\frac{\pi i}{6}\right)+\text{res}\left(f,e^\frac{\pi i}{2}\right) +\text{res}\left(f,e^\frac{5\pi i}{6}\right)\right) \\ &=& 2\pi i\left(\text{res}\left(f,e^\frac{\pi i}{6}\right)+\text{res}\left(f,i\right) +\text{res}\left(f,e^\frac{5\pi i}{6}\right)\right). \end{eqnarray*} {/eq}

The following result from complex analysis is used to calculate each residue.

Fact: Let {eq}g(z) {/eq} and {eq}h(z) {/eq} be analytic at {eq}z_0 {/eq}. If {eq}g(z_0)\neq0 {/eq}, {eq}h(z_0)=0 {/eq} and {eq}h'(z_0)\neq0 {/eq} then

{eq}\begin{eqnarray*} \text{res}(f,z_0) &=& \frac{g(z_0)}{h'(z_0)}. \end{eqnarray*} {/eq}

In this case, {eq}f(z)=\frac{g(z)}{h(z)} {/eq} where {eq}g(z)=1 {/eq} and {eq}h(z)=z^6+1 {/eq}. Now {eq}h'(z)=6z^5 {/eq}. At each pole {eq}z_k {/eq} for {eq}k=0,1,2 {/eq} we clearly have {eq}g(z_k)=1\neq0 {/eq}, {eq}h(z_k)=0 {/eq} and {eq}h'(z_k)\neq0 {/eq}.

By the fact we have

{eq}\begin{eqnarray*} \text{res}\left(f,e^\frac{\pi i}{6}\right) &=& \frac{1}{6\left(e^\frac{\pi i}{6}\right)^5} \\ &=& \frac{1}{6}\left(e^{-\frac{5\pi i}{6}}\right) \\ \text{res}(f,i) &=& \frac{1}{6i^5} \\ &=& \frac{1}{6i} \\ \text{res}\left(f,e^\frac{5\pi i}{6}\right) &=& \frac{1}{6\left(e^\frac{5\pi i}{6}\right)^5} \\ &=& \frac{1}{6\left(e^\frac{25\pi i}{6}\right)} \\ &=& \frac{1}{6\left(e^\frac{24\pi i}{6}e^\frac{\pi i}{6}\right)} \\ &=& \frac{1}{6\left(e^{4\pi i}e^\frac{\pi i}{6}\right)} \\ &=& \frac{1}{6\left(e^\frac{\pi i}{6}\right)} \\ &=& \frac{1}{6}\left(e^{-\frac{\pi i}{6}}\right). \end{eqnarray*} {/eq}

By the residue formula we get

{eq}\begin{eqnarray*} I &=& \int_{x=-\infty}^\infty\frac{1}{x^6+1}~dx \\ &=& 2\pi i~\left(\frac{1}{6}\left(e^{-\frac{5\pi i}{6}}\right) +\frac{1}{6i}+\frac{1}{6}\left(e^{-\frac{\pi i}{6}}\right)\right) \\ &=& \frac{2\pi i}{6}\left(e^{-\frac{5\pi i}{6}}+\frac{1}{i}+e^{-\frac{\pi i}{6}}\right) \\ &=& \frac{\pi i}{3}\left(\cos\left(-\frac{5\pi}{6}\right)+i\sin\left(-\frac{5\pi}{6}\right)+\frac{1}{i} +\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right) \\ &=& \frac{\pi i}{3}\left(\cos\left(\frac{5\pi}{6}\right)-i\sin\left(\frac{5\pi}{6}\right)+\frac{1}{i} +\cos\left(\frac{\pi}{6}\right)-i\sin\left(\frac{\pi}{6}\right)\right) \\ &=& \frac{\pi i}{3}\left(-\cos\left(\frac{\pi}{6}\right)-i\sin\left(\frac{\pi}{6}\right)+\frac{1}{i} +\cos\left(\frac{\pi}{6}\right)-i\sin\left(\frac{\pi}{6}\right)\right) \\ &=& \frac{\pi i}{3}\left(\frac{1}{i}-2i\sin\left(\frac{\pi}{6}\right)\right) \\ &=& \frac{\pi i}{3}\left(\frac{1}{i}-2i\left(\frac{1}{2}\right)\right) \\ &=& \frac{\pi i}{3}\left(\frac{1}{i}-i\right) \\ &=& \frac{\pi}{3}+\frac{\pi}{3} \\ &=& \frac{2\pi}{3}. \end{eqnarray*} {/eq}

Since {eq}f(x)=\frac{1}{x^6+1} {/eq} is an even function

{eq}\begin{eqnarray*} I &=& \int_{x=-\infty}^\infty\frac{1}{x^6+1}~dx \\ &=& 2\int_{x=0}^\infty\frac{1}{x^6+1}~dx \\ &=& 2I_1. \end{eqnarray*} {/eq}

Therefore

{eq}\begin{eqnarray*} \int_{x=0}^\infty\frac{1}{x^6+1}~dx &=& I_1 \\ &=& \frac{1}{2}I \\ &=& \frac{1}{2}\left(\frac{2\pi}{3}\right) \\ &=& \frac{\pi}{3}. \end{eqnarray*} {/eq}


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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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