Integrate $\int_{0}^{\infty} \frac{x \sin(x)}{(1 + x^{4})} dx$
I'm having some trouble with the integral on the title.
I noticed that it is an even function, so it is the same as doing $\frac{1}{2}\int_{-\infty}^{\infty}\frac{x\sin(x)}{(1 + x^{4})} dx$
It has 4 simple poles which are z$_{1}$=e$^{\frac{\pi i}{4}}$, z$_{2}$=e$^{\frac{3\pi i}{4}}$, z$_{3}$=e$^{\frac{5\pi i}{4}}$, z$_{4}$=e$^{\frac{7\pi i}{4}}$
I tried using the residue theorem to calculate it as
I = $\pi$i$\cdot$[Res$_{z=e^{\frac{\pi i}{4}}}$f(z) + Res$_{z=e^{\frac{3\pi i}{4}}}$f(z)]
and I get
I = $\frac{\pi}{2}\cdot \sin(\frac{1}{\sqrt{2}})\cdot\frac{1 + e^{\sqrt{2}}}{2e^{\frac{\sqrt{2}}{2}}}$
but when I put it on WolframAlpha to check my result I get
I = $\frac{\pi}{2}\cdot \sin(\frac{1}{\sqrt{2}})\cdot e^{\frac{-1}{\sqrt{2}}}$
Help, I don't know what I'm doing wrong. I'm pretty sure I'm calculating the residue right, but maybe I can't use that theorem?
$\endgroup$ 31 Answer
$\begingroup$Your integral is$$\frac12\Im\int_{\Bbb R}\frac{xe^{ix}dx}{1+x^4}.$$The poles are at $\frac{\pm1\pm i}{\sqrt{2}}$, all first-order. However, an infinite contour that enclosed a pole of negative imaginary part would enclose points with $\Im z\to-\infty$ i.e. $|e^{iz}|\to\infty$, so such poles must be ignored. Therefore, we calculate the integral from the residues at $z_\pm:=\frac{\pm1+i}{\sqrt{2}}$. Now$$\lim_{z\to z_\pm}\frac{z(z-z_\pm)e^{iz}}{1+z^4}=\frac{e^{iz_\pm}}{4z_\pm^2}=\frac{e^{(-1\pm i)/\sqrt{2}}}{\pm 4i}.$$So the original integral is$$\frac{\pi}{2}\Im e^{(-1+i)/\sqrt{2}}=\frac{\pi}{2}e^{-1/\sqrt{2}}\sin\frac{1}{\sqrt{2}}.$$
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