Need help calculating a Residue

**kevinlightman** · December 24th 2010, 08:57 AM

I am trying to integrate log(x)/( (1+x^2)^2) and I'm stuck at trying to find the residue of that function at i. I've tried expanding it to find the coefficient of (x-i) but I just get into a mess and as the pole is a double covert pole I don't think there is any nice formula that would lead me to the answer. Can anyone help please?

**Drexel28** · December 24th 2010, 09:40 AM

Originally Posted by kevinlightman

I am trying to integrate log(x)/( (1+x^2)^2) and I'm stuck at trying to find the residue of that function at i. I've tried expanding it to find the coefficient of (x-i) but I just get into a mess and as the pole is a double covert pole I don't think there is any nice formula that would lead me to the answer. Can anyone help please?

Let $\displaystyle f(z)=\frac{\log(z)}{(1+z^2)^2}$ . Then, you noted that $f$ has pole of order $2$ at $i$ so that

$\displaystyle \begin{aligned}\underset{z=i}{\text{Res }}f(z) &= \lim_{z\to i}\frac{d}{dz}\left((z-i)^2f(z)\right)\\ &= \lim_{z\to i}\frac{z-2z\log(z)+i}{z(z+i)^3}\\ &= \frac{\pi+2i}{8}\end{aligned}$

**FernandoRevilla** · December 24th 2010, 10:07 AM

Originally Posted by kevinlightman

I am trying to integrate log(x)/( (1+x^2)^2)

Perhaps you meant:

$I=\displaystyle\int_0^{+\infty}\dfrac{\log x\;dx}{(1+x^2)^2}$

Use Drexel28's post to prove:

$I=\ldots=\dfrac{\pi\log 2}{4}$

Fernando Revilla

**Drexel28** · December 24th 2010, 10:38 AM

Originally Posted by FernandoRevilla

Perhaps you meant:

$I=\displaystyle\int_0^{+\infty}\dfrac{\log x\;dx}{(1+x^2)^2}$

Use Drexel28's post to prove:
$I=\ldots=\dfrac{\pi\log 2}{4}$

Fernando Revilla

Indeed, one can use the semi-circle contour $\Gamma_R=\left\{z:|z|=R\right\}\cap \text{URHP}$ and use the Residue THeorem.

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