8
$\begingroup$

how do I compute

$$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx$$

with complex analysis?

I feel like im calculating the residue wrong and I cant get to the answer correctly. I tried to branch cut the real $0 \rightarrow \infty$ but I feel like im doing it wrong. any help is appriciated.

additional information:

thank you for the input everyone it is very helpful.

i did come down to calculating the integral

$$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx = i\pi [Res(f,z_1=-1+2i)+Res(f,z_2=-1-2i)]$$

Then given answer to this question is $\frac{\pi}{2}\sqrt{\frac{\sqrt{5}-1}{2}}$

I was just simply calculating

$i\pi [Res(f,z=-1+2i)+Res(f,z=-1-2i)] = i\pi \left(\frac{\sqrt{z_1}}{2z_1+2}+\frac{\sqrt{z_2}}{2z_2+2}\right)$

Solving for

$\frac{\pi}{4} \left(\sqrt{-1+2i}-\sqrt{-1-2i}\right)$

I get $\frac{\pi}{2}\sqrt{\frac{-\sqrt{5}-1}{2}}$ and I still dont know what I am doing wrong for that one sign error.

$\endgroup$
  • 1
    $\begingroup$ That's the right branch-cut. Can you add what you did? It's hard to spot any mistakes without seeing that. $\endgroup$ – Daniel Fischer Nov 19 '14 at 17:21
  • $\begingroup$ Perhaps you computed the wrong value for $\sqrt{-1-2i}$. With the branch cut given, it is $-\sqrt{\phi-1}+i\sqrt\phi$ and not $\sqrt{\phi-1}-i\sqrt\phi$. $\endgroup$ – robjohn Nov 19 '14 at 18:53
12
$\begingroup$

The integral along the contour just above the real axis would be $$ \int_0^R\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x\tag{1} $$ The integral along the contour circling the complex plane at a radius of $R$ would be bound by $$ \begin{align} \int_0^{2\pi}\frac{\sqrt{R}}{R^2-2R-5}R\,\mathrm{d}\theta &\stackrel{\hphantom{R\to\infty}}{\le}\frac{2\pi}{R^{1/2}-2R^{-1/2}-5R^{-3/2}}\\ &\stackrel{R\to\infty}{\to}0\tag{2} \end{align} $$ The integral along the contour just below the real axis would be $$ \int_R^0\frac{-\sqrt{x}}{x^2+2x+5}\mathrm{d}x\tag{3} $$ Adding up the pieces $(1)$, $(2)$, and $(3)$, and letting $R\to\infty$, we get $$ \begin{align} 2\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x &=2\pi i\left[\vphantom{\frac{\sqrt{-1-2i}}{-4i}}\right.\underbrace{\frac{\sqrt{-1-2i}}{-4i}}_{\begin{array}{}\text{Residue at}\\z=-1-2i\end{array}}+\underbrace{\frac{\sqrt{-1+2i}}{4i}}_{\begin{array}{}\text{Residue at}\\z=-1+2i\end{array}}\left.\vphantom{\frac{\sqrt{-1-2i}}{-4i}}\right]\\ &=2\pi i\left[\frac{-\sqrt{\phi-1}+i\sqrt\phi}{-4i}+\frac{\sqrt{\phi-1}+i\sqrt\phi}{4i}\right]\\[12pt] &=\frac\pi{\sqrt\phi}\tag{4} \end{align} $$ and therefore, $$ \int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi}\tag{5} $$

$\endgroup$
  • $\begingroup$ From what I've been taught, you also have to consider a second circle around the branch point whose radius goes to $0$. This integral does go to $0$, it is nontrivial $\endgroup$ – Dylan Nov 28 '17 at 12:37
  • $\begingroup$ @Dylan: the integrand near $x=0$ vanishes, so the integral around an infinitesimal semi-circle will be the integral of a vanishing function over an infinitesimal contour; so it will be $0$. I didn't think it was really worth mentioning, but perhaps I will move part of this comment into the answer. $\endgroup$ – robjohn Nov 28 '17 at 14:33
  • $\begingroup$ Practically it's just a way to close the contour, since the two line segments are separated by some distance in the absence of a limit. I'll say it's definitely less important than the bigger circle though. $\endgroup$ – Dylan Nov 28 '17 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.