I am asked to calculate a complex integral. how to compute $\displaystyle \int \limits_{-\infty}^{\infty}\frac{x^4}{1+x^8}dx$ with residue theorem?
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What's troubling you? – Git Gud Jul 22 '14 at 14:27
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compute the residues – user133834 Jul 22 '14 at 14:28
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1See this. – Tunk-Fey Jul 22 '14 at 14:30
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This doesn't look like a duplicate to me. The other question doesn't specify that it is to be done by using the residue theorem. – Michael Hardy Jul 22 '14 at 15:56
2 Answers
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Some hints that might help you along the way:
Start off by writing the expression as $$ \oint_{C} \frac{z^4}{1 + z^8}dz $$ where $C$ is the contour. Now find the residues: $$ 1 + z_n^8 = 0, z_n \in C $$ Then the integral value is $$ 2 \pi i \sum Res_{z=z_n}f(z) $$
Noxet
- 760
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You might be interested in an alternative method using the Beta function. Note that the integrand is even, so
\begin{align}
\int_{-\infty}^{\infty}\frac{x^4}{1+x^8}dx
&=2\int_{0}^{\infty}\frac{x^4}{1+x^8}dx\\
\end{align}
Use the substitution $x=u^{1/8}$, then $dx=\frac{1}{8}u^{-7/8}du$. The integral becomes
\begin{align}
2\int_{0}^{\infty}\frac{x^4}{1+x^8}dx
&=\frac{1}{4}{\int^{\infty}_{0}\frac{u^{-3/8}}{1+u}du}\\
&=\frac{1}{4}\int^{1}_{0}\frac{(\frac{y}{1-y})^{-3/8}}{1+\frac{y}{1-y}}\frac{dy}{(1-y)^2}\\
&=\frac{1}{4}\int^{1}_{0}y^{-3/8}(1-y)^{-5/8}dy\\
&=\frac{1}{4}B(5/8,\ 3/8)\\
&=\frac{\Gamma(5/8)\ \Gamma(3/8)}{4\ \Gamma(1)}\\
&=\frac{\pi}{4}\csc{\frac{3\pi}{8}}
\end{align}
I have made use of the definition of the Beta function and Euler's Reflection Formula.
If you prefer a complex analysis approach, the poles are at
\begin{align}
z=e^{i (2n+1)\pi /8}, \ n\in\mathbb{Z}, \ 0\le n \le 7
\end{align}
Figure out which singularities lie in your contour and proceed with the integration.
SuperAbound
- 5,554