It's a very simple problem. I put it into wolfram alpha and it gave me a hugely complicated answer. This makes me think that I am supposed to be doing it differently.

I believe I am supposed to use the residue theorem for this answer but have no idea how. $$\int_{0}^{\infty}\frac{x^2}{x^4+16}dx = 2\pi i \sum_{n=0}^{\infty} Res\big(\frac{x^2}{x^4+16}\big)$$ I have no idea what Res() is or how to use this in any way.

Any and all help is appreciated, thank you.

Sorry, I typed the equation in wrong, I just corrected it from $\frac{x^4}{x^2+16}$ to $\frac{x^2}{x^4+16}$

  • $\begingroup$ Why not $x=4\tan t$ $\endgroup$ – lab bhattacharjee Oct 19 '18 at 4:28
  • 2
    $\begingroup$ Are you sure that's not supposed to be $x^2/(x^4 + 16)$? $\endgroup$ – eyeballfrog Oct 19 '18 at 4:35
  • $\begingroup$ @eyeballfrog is correct, I have since updated the equation, thank you $\endgroup$ – Big_Tubbz Oct 19 '18 at 4:38
  • $\begingroup$ Have you not taken a course in complex analysis? That's were residues are introduced, explained and utilized in evaluating integrals like this. Presumably that course is a prerequisite here? Do observe that you only need finitely many residues here. That infinite sum looks strange. $\endgroup$ – Jyrki Lahtonen Oct 19 '18 at 5:01
  • $\begingroup$ @JyrkiLahtonen I have not. This is for an introduction to mathematical physics course and is the first time I have seen the residue theorem. The only prerequisites are calculus, linear algebra, and quantum/relativistic physics. $\endgroup$ – Big_Tubbz Oct 19 '18 at 5:13

If you call the integral $I$, then $$2I=\int_{-\infty}^\infty\frac{z^2\,dz}{z^4+16}.$$ This can be done by the "semicircle" method in contour integration, and equals $2\pi i$ times the sum of the residues of the integrand of the poles in the upper half plane. These poles are at $2\zeta$ and $2\zeta^3$ where $\zeta=\exp(\pi i/4)$. The residue at $2\zeta$ is $$\lim_{z\to2\zeta}\frac{(z-2\zeta)z^2}{z^4+16}=\frac{(2\zeta)^2} {4(2\zeta)^3}=\frac{1}{8\zeta}.$$ Now do the very similar calculation for the other residue and put the pieces together.

  • $\begingroup$ I am having trouble following the step in removing the limit, does the $(z-2\zeta)$ factor out from $(z^{4} + 16)$? $\endgroup$ – Big_Tubbz Oct 19 '18 at 5:14
  • 1
    $\begingroup$ It does indeed, but it's easier to consider the derivative of $z^4+16$ at $z=2\zeta$. @Big_Tubbz $\endgroup$ – Lord Shark the Unknown Oct 19 '18 at 5:17
  • $\begingroup$ Thank you for everything so far, I have found a way to solve the equation, but I keep getting the wrong answer. is the residue at $2 \zeta = \frac{1}{8 \zeta^{3}}$? $\endgroup$ – Big_Tubbz Oct 19 '18 at 5:46
  • $\begingroup$ Sorry, I mean the residue at $2 \zeta^{3}$ $\endgroup$ – Big_Tubbz Oct 19 '18 at 5:55
  • $\begingroup$ @LordTheShark I am the downvote, and it was accidental. If you edit your answer I can fix it, but for now it’s been too long since my vote so I cannot change it. $\endgroup$ – JavaMan Oct 19 '18 at 14:20

With substitution $x^2=4\tan u$ then \begin{align} \int_0^\infty\dfrac{x^2}{x^4+16}\ dx &= \dfrac14\int_0^{\pi/2}\sqrt{\tan u} \ du \\ &= \dfrac14\int_0^{\pi/2}\sin^{2(\frac34)-1}u \cos^{2(\frac14)-1}u\ du \\ &= \dfrac18\beta(\dfrac34,\dfrac14) \\ &= \dfrac18\Gamma(\dfrac34)\Gamma(\dfrac14) \\ &= \color{blue}{\dfrac{\pi}{4\sqrt{2}}} \end{align} here $\beta(x,y)$ is Beta function and $$\Gamma(1+x)\Gamma(1-x)=\dfrac{\pi x}{\sin\pi x}$$

  • $\begingroup$ What is $β$ and what are the $Γ$? $\endgroup$ – Big_Tubbz Oct 19 '18 at 4:50
  • $\begingroup$ The beta and gamma functions. But the more frequent notation for the beta function is $B(x,y)$ rather than $\beta(x,y)$. $\endgroup$ – Lord Shark the Unknown Oct 19 '18 at 4:54

I will provide another complex method.
Let $f(z)=\frac{z^2\ln (-z)}{z^4+16}$. Taking $C$ as the keyhole contour (small circle around the origin, upward of positive real axis, big circle, downward of positive real axis). $$\int_C f(z)dz=2\pi i\sum_{z:\text{roots of $z^4+16$}}\operatorname{Res}(f(z))=2\pi i(-\frac{\pi}{4\sqrt2}),$$ $$I=\frac{2\pi i}{-2\pi i}(-\frac{\pi}{4\sqrt2})=\frac{\pi}{4\sqrt2}$$


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