2

Calculate: $$\int_{0}^{\infty}\frac{\sin x}{x^{3}+x}\mathrm{d}x$$
My try:

Let's split to a pacman style path, with little circle around the singularity and 2 rays from $0$ to $\infty$:

$\displaystyle \int_{0}^{\infty}\frac{\sin x}{x^{3}+x}\mathrm{d}x=\Im\int_{|z|=\varepsilon}\frac{\varepsilon e^{\theta i}}{(\varepsilon e^{\theta i})^{3}+\varepsilon e^{\theta i}}\mathrm{d}\theta+\Im\int_{0}^{\infty}\frac{re^{\delta i}}{(re^{\delta i})^{3}+re^{\delta i}}\mathrm{d}r+\Im\int_{0}^{\infty}\frac{re^{-\delta i}}{(re^{-\delta i})^{3}+re^{\delta i}}\mathrm{d}r$

where $\varepsilon \to 0$, and then the integral is $1$ therefore it's imaginary part is $0$.

$\delta$ is again an angle as small as we want (as it need not contain the pole at $\pm i$. According to the residue theorem the domain contains no poles therefore their sum must be zero, therefore the whole integral must be $0$, but this is not the correct answer. What claim I make is wrong?

V.G
  • 4,196
hash man
  • 129
  • Remember that $r \to +\infty$ and $\varepsilon \to 0$... So, for relevant values of $r,\varepsilon$, there will be a pole inside the contour. – PierreCarre Aug 06 '20 at 10:57
  • if I take epsilon small enough and delta small enough there won't be any – hash man Aug 06 '20 at 10:58
  • for epsilon small enough the integral becomes the integral on 1. so I can't understand your statement. – hash man Aug 06 '20 at 11:00
  • I don't know what you call $\delta$... the proper integration curve would be the upper semi-circle of radius $r>1$, with a small blob (semi-circle of radius $\varepsilon$) to avoid having $z=0$ on the curve. To this integral you apply the residue theorem and then see what happens when $\varepsilon \to 0$ and $r \to +\infty$. – PierreCarre Aug 06 '20 at 15:40
  • There are several issues: the change of variables, the orientation of the parts of the contour and the assumption that the lhs and the rhs are equal. You can find the integral of $e^{i x}/(x^3 + x)$ over $|x| = r$ (which has finite limits when $r \to 0^+$ and when $r \to \infty$), but it doesn't directly give the value of the original integral $I$. You can however use $$2 I = \operatorname {Im} \operatorname {v. ! p.} \int_{\mathbb R} \frac {e^{i x}} {x^3 + x} dx$$ or $$2 I = \int_{-\infty +i0}^{\infty + i0} \frac {e^{i x} - e^{-i x}} {2 i (x^3 + x)} dx.$$ – Maxim Aug 09 '20 at 02:36

2 Answers2

1

You're missing the outer circle to close the loop, which diverges. Furthermore, the rays around the positive real line will cancel out, so the given integral does not simply equal the integrals you wrote.

A better integral to use would be a semicircle contour avoiding the origin. Note also that the integrand is symmetric:

$$\int_0^\infty\frac{\sin(x)}{x^3+x}~\mathrm dx=\frac12\int_{-\infty}^\infty\frac{\sin(x)}{x^3+x}~\mathrm dx$$

The inner (clockwise semi-) circle will converge to $-\pi i$ while the outer circle will vanish, leaving us with

$$-\pi i+\int_{-\infty}^\infty\frac{e^{iz}}{z^3+z}~\mathrm dz=\oint_C\frac{e^{iz}}{z^3+z}~\mathrm dz=2\pi i\underset{z=i}{\operatorname{Res}}\frac{e^{iz}}{z^3+z}$$

and you should be able to take it from here.

Simply Beautiful Art
  • 74,685
0

Another approach that doesn't requires the use of complex analysis would be:

We have $$\int_{0}^{\infty}\frac{\sin x}{x^3+x}\mathrm{d}x=\int_{0}^{\infty}\left(\frac{\sin x}{x}-\frac{x\sin x}{x^2+1}\right)\mathrm{d}x=\underbrace{\int_{0}^{\infty}\frac{\sin x}{x}\mathrm{d}x}_{I_1}-\underbrace{\int_{0}^{\infty}\frac{x\sin x}{x^2+1}\mathrm{d}x}_{I_2} $$

$I_1$ is well known to be equal to $\dfrac{\pi}{2}$ and solved here and $I_2$ has been solved here using the method of parameterization/Feynman's technique and is found to be $\dfrac{\pi}{2e}$.

Thus our answer is $\boxed{\frac{(e -1)\pi}{2e}}$.

V.G
  • 4,196