I'm looking for a better approach to compute $\int_{0}^{\infty} \frac{x\sin(\pi x)}{(x^2+4)^3} dx$.

Our contour $\mathcal C$ consists of a upper half-plane semicircle and the real axis joining the ends, where we take the limit of the radius to infinity.

By Jordan's lemma, the semicircular part of $\mathcal C$ vanishes (first equality). Which leaves us to use the residue theorem (second equality) to compute,

$$ \require{cancel} I := \int_{0}^{\infty} \frac{x\sin(\pi x)}{(x^2+4)^3}dx = \frac{1}{2}\oint_{\mathcal C}f(z)dz = \pi \big(R\{f,2i\}\cancel{+R\{f,-2i\} }\big)\ . $$

As both poles are the pole is of third order, we need to take two a second derivative

$$ \lim_{z\to 2i} \frac{1}{2!} \frac{d^2}{dz^2}\bigg( (z+2i)^{-3}z\sin(\pi z) \bigg) $$ in order to find the residue at $z=2i$.

Although this seems tedious, we'll give it a try.

To simplify the notation, put $a:=(z+2i)^{-1}$, $\ b:= \pi \cos(\pi z)\ $ and $\ c:= \sin(\pi z).$

For the first derivative on the left, this yields

$$ (-3a^4 z + a^3)c + a^3bz \ , $$ which we can reorganize as

$$ -3a^4 zc + a^3(c + bz) \ . $$

The second derivative yields

$$ 12a^5 cz-3a^4(c+bz) -3a^4(c+bz) + a^3(b-\pi^2 cz + b) , $$

which we can reorganize as

$$ 12a^5 cz -6a^4(c+bz) + a^3(2b-\pi^2 cz) \\ = 12(z+2i)^{-5}z\sin(\pi z) -6(z+2i)^{-4}\big(\sin(\pi z) + \pi z \cos(\pi z)\big) +(z+2i)^{-3}\big( \sin(\pi z) + \pi z \cos(\pi z) \big) . $$

Plugging in $z=2i$ yields (if I didn't make any calculation mistakes)

$$ R\{f,2i\} = -2^{-7}\big(\sin(2i \pi)(2\pi^2 -3) + i \cos(2i\pi) \big) \ . $$

Wolfram Alpha says the result should be $I = 2^{-7}e^{-2\pi} \pi^2 (1+2\pi)$.

Even ìf this approach turns out to work, short of a few calculations mistakes, I don't find it particularly efficient.

I'm wondering whether there is a more practical approach for computing the residue $R\{f,2i\}$, or even $I$.

  • 4
    $\begingroup$ do an integration by parts: $$ ´-\frac{1}{4}\int_0^{\infty}\frac{\cos(\pi x)}{(x^2+4)^2}dx $$ is a lot nicer. The order of the pole is reduced as well as the complexity of the integrand $\endgroup$
    – tired
    Jan 26, 2017 at 14:30
  • 1
    $\begingroup$ Why are you adding in the residue at $-2i$? The curve doesn't surround that singularity. $\endgroup$ Jan 26, 2017 at 14:46
  • $\begingroup$ @AdamHughes Yes, you're right. Thanks for pointing it out. :) I'll make an edit. $\endgroup$
    – easytarget
    Jan 26, 2017 at 14:48
  • $\begingroup$ @tired I think you're missing a factor $\pi$. $\endgroup$
    – easytarget
    Jan 26, 2017 at 15:23
  • $\begingroup$ @MusséRedi yeah, but this doesn't change anything conceptually ^^ $\endgroup$
    – tired
    Jan 26, 2017 at 15:24

1 Answer 1


If we perform a preliminary step of integration by parts as (wisely) suggested by tired, we get

$$ I = \frac{\pi}{8}\int_{-\infty}^{+\infty}\frac{\cos(\pi x)}{(x^2+4)^2}\,dx = \frac{2\pi^2 i}{8}\,\text{Res}\left(\frac{e^{\pi i x}}{(x^2+4)^2},x=2i\right) $$ where $$\text{Res}\left(\frac{e^{\pi i x}}{(x^2+4)^2},x=2i\right)=\lim_{x\to 2i}\frac{d}{dx}\frac{e^{\pi i x}}{(x+2i)^2} =-\frac{i}{32}(2\pi+1)e^{-2\pi}$$ leads to $I =\color{red}{\large \frac{\pi^2(2\pi+1)}{128 e^{2\pi}}}$ without too much effort.

  • $\begingroup$ In taking the residue of $\frac{e^ {\pi i x}}{(x^2+4)^2}$ instead of $\frac{\cos(\pi x)}{(x^2+4)^2}$ are we using the oddness of the sin function; noticing that the sin part would vanish in the first integral? Because, I'm wondering why $ \operatorname{Res} \bigg( \frac{\cos(\pi x)}{(x^2+4)^2},x=2i \bigg)$ isn't giving the same answer for me. $\endgroup$
    – easytarget
    Jan 26, 2017 at 19:32
  • 1
    $\begingroup$ Never mind. We arrive at the result by equating the real part of both sides $$ \int_{-\infty}^{+\infty} \frac{e^{i\pi x}}{(x^2+4)^2} dx = 2\pi i \operatorname{Res} \bigg( \frac{e^{i\pi x}}{(x^2+4)^2}, x =2i \bigg) \ . $$ $\endgroup$
    – easytarget
    Jan 29, 2017 at 11:06

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