6
$\begingroup$

Using residue calculus show that $$\int_0^{\infty}\frac{\log(x^2+4)}{(x^2+1)^2}dx=\frac{\pi}2\log 3-\frac{\pi}6.$$

I was thinking of using some keyhole or semi-circular contour here. But the problem is apart from poles at $x=-i$ and $x=i$, the logarithm has singularities when $x=\pm 4i$.

I consider $C_R$, a semicircle contour oriented clockwise with radius $R$ centered at origin. The semicircle resides in the lower half plane, so that it encloses $x=-i$ as a pole. I set $$\int_{C_R} \frac{\log(2-ix)}{(x^2+1)^2}dx$$ But it seems like it leads to wrong answer.

$\endgroup$
1
  • $\begingroup$ You moving in the right direction. Please note that the you have a second order pole (if you go clockwise - in the negative direction - and close the contour in the lower half-plane). $\int_{C_R} \frac{\log(2-ix)}{(x^2+1)^2}dx=-2\pi{i}Res\frac{\log(2-ix)}{(x^2+1)^2}|_{x=-i}=-2\pi{i}\frac{d}{dx}\frac{\log(2-ix)}{(x-i)^2}|_{x=-i}$. And then you have to take a real part of it and divide by 2 - to get the initial integral. $\endgroup$ – Svyatoslav Mar 7 at 12:45
6
$\begingroup$

Integrating $$ \int_\gamma\frac{\log\left(z^2+4\right)}{\left(z^2+1\right)^2}\,\mathrm{d}z\tag1 $$ along the contour

enter image description here

gives $2\pi i$ times the residue at $z=i$.

The integral along the curved pieces vanish as the radius of the large semi-circle grows to $\infty$.

The integral along the downward line along the right side of the imaginary axis to $2i$ (in red) is $$ -i\int_2^\infty\frac{\log\left(x^2-4\right)+\pi i}{\left(x^2-1\right)^2}\,\mathrm{d}x\tag2 $$ The integral along the upward line along the left side of the imaginary axis from $2i$ (in green) is $$ i\int_2^\infty\frac{\log\left(x^2-4\right)-\pi i}{\left(x^2-1\right)^2}\,\mathrm{d}x\tag3 $$ For $z=ix$ and $x\gt2$, we have one of $\log\left(z^2+4\right)=\log\left(x^2-4\right)\pm\pi i$. Integrating $\frac1{z-2i}+\frac1{z+2i}$ clockwise around $2i$ decreases $\log\left(z^2+4\right)$ by $2\pi i$. Thus, on the right side of the branch cut, we have $\log\left(x^2-4\right)+\pi i$, and on the left side, we have $\log\left(x^2-4\right)-\pi i$.

The sum of the integrals in $(2)$ and $(3)$ is $$ 2\pi\int_2^\infty\frac1{(x^2-1)^2}\,\mathrm{d}x=\frac\pi6(4-3\log(3))\tag4 $$ The integral along the real axis (in blue) is $$ 2\int_0^\infty\frac{\log(x^2+4)}{(x^2+1)^2}\,\mathrm{d}x\tag5 $$ Furthermore, $$ 2\pi i\operatorname*{Res}_{z=i}\left(\frac{\log(z^2+4)}{(z^2+1)^2}\right)=\frac\pi6(2+3\log(3))\tag6 $$ Subtracting $(4)$ from $(6)$ and dividing by $2$ gives $$ \int_0^\infty\frac{\log(x^2+4)}{(x^2+1)^2}\,\mathrm{d}x=\frac\pi2\log(3)-\frac\pi6\tag7 $$

$\endgroup$
2
  • $\begingroup$ Very nice presentation. (+1) $\endgroup$ – Markus Scheuer Mar 8 at 19:02
  • 1
    $\begingroup$ @MarkusScheuer: thanks! I have added a description of how $\log\left(z^2+4\right)$ behaves near the branch cut $\endgroup$ – robjohn Mar 8 at 21:31
1
$\begingroup$

One thing you could do to avoid a keyhole is to denote

$$I[a] = \int_0^\infty \frac{\log\left(a(x^2+1)+3\right)}{(x^2+1)^2}\:dx \implies I'[a] = \frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^2+1}\cdot \frac{1}{ax^2+a+3}\:dx$$

then apply a semicircular contour and calculate the residues as normal.

$\endgroup$

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.