I need help to solve the next improper integral using complex analysis:
∫∞−∞x6(4+x4)2dx
I have problems when I try to find residues for the function f=1(z4+4)2.
This is what I tried.
res(f,√2ei(π4+kπ2))=limz→√2ei(π4+kπ2)((z−√2ei(π4+kπ2))2(z4−4)2)′
with k\in\{0,1,2,3\}. What do you think about it?
I know there is a little more general problem involving this integral; for all a>0
\int_{-\infty}^{\infty} \frac{x^6}{(a^4+x^4)^2} dx= \frac{3\pi\sqrt{2}}{8a}
Edit.
I've had an idea: from the integration by parts
\int u dv = uv - \int vdu
and if we let
dv = \frac{4x^3 }{(4+x^4)^2}, \, u = \frac{x^3}{4}
with
dv = -\frac{d}{dx} \frac{1}{4+x^4} = \frac{4x^3 }{(4+x^4)^2}
we get finally
\int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx = 0 + \frac{3}{4} \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx
which I think is more easy to solve.
Anyway, if you know another idea or how to complete my first try will be welcome.
Answer
If you are right, then \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx can be easily done by a semi-circle contour, computing the residues at (1+i)/\sqrt{2} and (-1+i)/\sqrt{2}. It is easy to check that the integral of \frac{z^2}{1+z^4} on the arc of the semi-circle goes to 0.
\frac{z^2}{1+z^4}=\frac{z^2}{(z-e^{i\pi/4})(z-e^{i3\pi/4})(z-e^{i5\pi/4})(z-e^{i7\pi/4})} shows that it is a simple pole at those residues.
\text{ Res}(\frac{z^2}{1+z^4},e^{i\pi/4})=\lim_{z\to e^{i\pi/4}}\frac{z^2(z-e^{i\pi/4})}{1+z^4}=\lim_{z\to e^{i\pi/4}}\frac{3z^2-2ze^{i\pi/4}}{4z^3}=(\frac{3}{4}-\frac{1}{2})e^{-i\pi/4}.
\text{ Res}(\frac{z^2}{1+z^4},e^{i3\pi/4})=\lim_{z\to e^{i3\pi/4}}\frac{z^2(z-e^{i3\pi/4})}{1+z^4}=\lim_{z\to e^{i3\pi/4}}\frac{3z^2-2ze^{i3\pi/4}}{4z^3}=(\frac{3}{4}-\frac{1}{2})e^{-i3\pi/4}.
I used L'Hospital's rule above because it seems simpler.
So the answer is \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx=2\pi i\frac{1}{4}\left(\frac{1-i}{\sqrt{2}}+\frac{-1-i}{\sqrt{2}}\right)=\frac{\sqrt{2}}2\pi.
Edit: Something's wrong with the integration by parts; we do not seem to get the right answer. It is a 4 in the bottom and not 1. But we can do a further substitution to get it right.
\int_{-\infty}^{\infty} \frac{x^2}{4+x^4} dx=\frac{1}{\sqrt{2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2}}\frac{(\frac{x}{\sqrt{2}})^2}{1+(\frac{x}{\sqrt{2}})^4} dx=\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx.
So the final answer is \int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx=\frac{3}{4}\int_{-\infty}^{\infty} \frac{x^2}{4+x^4} dx=\frac{3}{8}\pi. You also have to argue that you are limiting r\to\infty in \left[\frac{x^3}{4(4+x^4)}\right]_{-r}^r which is why it is 0.
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