# Evaluating $\int\limits_0^\infty \frac{\log x} {(1+x^2)^2} dx$ with residue theory

I need a little help with this question, please!

I have to evaluate the real convergent improper integrals using RESIDUE THEORY (vital that I use this), using the following contour:

$$\int\limits_0^\infty \frac{\log x} {(1+x^2)^2} dx$$

Using this contour:

$R>1$ and $r<1$

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I edited. Is the expression correct? –  Pedro Tamaroff Mar 7 '12 at 18:06
Why is it "vital" to use one particular tool? If it's a homework question, then please tag it appropriately. –  Henning Makholm Mar 7 '12 at 18:49

I'll give my humble idea to show the integral is $-\dfrac{\pi}{4}$.

With a change of variables ($x=e^u$) we have that

$$\mathcal{I}=\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } \int\limits_{ - \infty }^\infty {\frac{{u{e^u}}}{{{{\left( {1 + {e^{2u}}} \right)}^2}}}du}$$

We can write this as

$${\mathcal I} = \int\limits_{ - \infty }^\infty {\frac{{u{e^{ - u}}}}{{{{\left( {{e^{ - u}} + {e^u}} \right)}^2}}}du}$$

Putting $u=-v$ we have that

$${\mathcal I} = \int\limits_{ - \infty }^\infty {\frac{{u{e^{ - u}}}}{{{{\left( {{e^{ - u}} + {e^u}} \right)}^2}}}du} = -\int\limits_{ - \infty }^\infty {\frac{{v{e^v}}}{{{{\left( {{e^{ - v}} + {e^v}} \right)}^2}}}dv}$$

This means that

$$2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } \int\limits_{ - \infty }^\infty {\frac{{u\left( {{e^{ - u}} - {e^u}} \right)}}{{{{\left( {{e^u} + {e^{ - u}}} \right)}^2}}}du}$$

We can write this in terms of the hiperbolic functions, to get

$$2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{1}{2}\int\limits_{ - \infty }^\infty {\frac{{u\sinh u}}{{\cosh^2 u}}du}$$

Integration by parts gives ($(\operatorname{sech} u)'=-\dfrac{{\sinh u}}{{\cosh^2 u}}$)

$$- \int\limits_{ - \infty }^\infty {\frac{{\sinh udu}}{{{{\cosh }^2}u}}} = \left[ {u\operatorname{sech} u} \right]_{ - \infty }^\infty - \int\limits_{ - \infty }^\infty {\frac{{du}}{{\cosh u}}}$$

Finally, you can easily check that

$$\int\limits_{ - \infty }^\infty {\frac{{du}}{{\cosh u}}} = \pi$$

and that $u \operatorname{sech} u$ is odd so the first term in the RHS is zero. Thus

\eqalign{ & 2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{2} \cr & I = \int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{4} \cr}

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Yep it is a homework question, sorry for the confusion –  Bany Mar 7 '12 at 18:54
@Bany Have you come up with any ideas on how to do it? –  Pedro Tamaroff Mar 7 '12 at 19:02
yea well the first step is to caclulate the reside, ive worked out that this is 1/i since the poles are (+/- i) of order 2; but only +i lies in the contour. now i need to work out limits of integrals but im lost at this part –  Bany Mar 7 '12 at 19:08

It is very interesting that this problem can be solved just by a few line using substitution without using residue. Let the integral be $I$ and $$J(a)=-\frac{1}{2}\int_0^\infty\frac{\log x}{x^2+a^2}dx.$$ Clearly $I=-J'(1)$. Using $x=au$, one has \begin{eqnarray} I(a)&=&\frac{1}{2a}\int_0^\infty\frac{\log a+\log x}{x^2+1}dx\\ &=&\frac{\log a}{2a}\int_0^\infty\frac{1}{x^2+1}dx+\frac{1}{2a}\int_0^\infty\frac{\log x}{x^2+1}dx\\ &=&\frac{\pi\log a}{4a}. \end{eqnarray} Now $$I=\frac{a}{da}\frac{\pi\log a}{4a}\bigg|_{a=1}=-\frac{\pi}{4}.$$ Here we used the fact that $$\int_0^\infty\frac{\log x}{x^2+1}dx=0.$$

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