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:

enter image description here

$R>1$ and $r<1$

  • $\begingroup$ I edited. Is the expression correct? $\endgroup$ – Pedro Tamaroff Mar 7 '12 at 18:06
  • $\begingroup$ @Bany I gave an answer using your contour, though I'm a bit late. I hope it helps! $\endgroup$ – Brightsun Jan 15 '16 at 15:06

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

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

  • $\begingroup$ Yep it is a homework question, sorry for the confusion $\endgroup$ – Bany Mar 7 '12 at 18:54
  • $\begingroup$ @Bany Have you come up with any ideas on how to do it? $\endgroup$ – Pedro Tamaroff Mar 7 '12 at 19:02
  • $\begingroup$ 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 $\endgroup$ – Bany Mar 7 '12 at 19:08

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

  • $\begingroup$ +1: Very clever! I edited the post a bit to make it clearer, since this method deserves more visibility, in my opinion! $\endgroup$ – Brightsun Jan 15 '16 at 15:17

Calling $\sigma$ the contour you drew, i.e. a big arc of radius $R$, a small arc of radius $r$ connected by two segments, let the angle subtended by the arcs approach $2\pi$ from below, so that the two segments are just above and below the positive real axis. Choosing the branch cut of the logarithm on the positive real axis, by the residue theorem we have $$ \oint_\sigma \frac{\ln^2 z}{(1+z^2)^2}dz=i2\pi \left(\text{Res}\frac{\ln^2z}{(1+z^2)^2}\Big|_{z=e^{i\pi/2}}+\text{Res}\frac{\ln^2z}{(1+z^2)^2}\Big|_{z=e^{i3\pi/2}}\right); $$ note that we wrote $i={e^{i\pi/2}}$ and $-i=e^{i3\pi/2}$ according to the chosen branch cut. Using the formula for the second order pole residue: $$ f(z) = \frac{\ln^2z}{(1+z^2)^2}=\frac{\ln^2z}{(z+i)^2(z-i)^2} $$ $$ \frac{d}{dz}\left[f(z)(z-i)^2\right]=2\ln z\frac{(z+i)-z\ln z}{z(z+i)^3}=\Big|_{z=e^{i\pi/2}}i\frac{\pi}{8}\left(\frac{\pi}{2}+2i\right) $$ $$ \frac{d}{dz}\left[f(z)(z+i)^2\right]=2\ln z\frac{(z-i)-z\ln z}{z(z-i)^3} =\Big|_{z=e^{i3\pi/2}}-i\frac{3\pi}{8}\left(\frac{3\pi}{2}+2i\right) $$ so that $$ \oint_\sigma \frac{\ln^2 z}{(1+z^2)^2}dz=i2\pi\left[ i\frac{\pi}{8}\left(\frac{\pi}{2}+2i\right) -i\frac{3\pi}{8}\left(\frac{3\pi}{2}+2i\right) \right]=\pi^3+i\pi^2. $$

Now, the integral on the big circle vanishes as $R\to\infty$, since the integrand falls off as $\ln^2 R/(1/R^2)^2$, and the integral on the small circle vanishes as $r\to0$ since the $r$ coming from the line element cancels the $\ln^2r$ singularity: explicitly $$ \left|\int_\Gamma \frac{\ln^2z}{(1+z^2)^2}dz\right|\le\int_{0}^{2\pi}\frac{|\ln R + i\varphi|^2}{|1+R^2|^2}Rd\varphi\le\text{const}\times R^{-3}\xrightarrow[R\to\infty]{}0, $$ $$ \int_\gamma \frac{\ln^2z}{(1+z^2)^2}dz= \int_{0}^{2\pi}\frac{(\ln r + i\varphi)^2}{(1+re^{i2\pi})^2}ie^{i\varphi}rd\varphi\xrightarrow[r\to0]{}0. $$ Therefore, in such limits, $$ \pi^3+i\pi^2 = \int_0^{+\infty}\frac{\ln^2x}{(1+x^2)^2}dx-\int_{0}^{+\infty}\frac{(\ln x+i2\pi)^2}{(1+x^2)^2}dx \\= -i4\pi\int_{0}^{+\infty}\frac{\ln x}{(1+x^2)^2}dx+4\pi^2\int_{0}^{+\infty}\frac{1}{(1+x^2)^2}dx. $$ Comparing real and imaginary parts \begin{align} \int_{0}^{+\infty}\frac{dx}{(1+x^2)^2} &= \frac{\pi}{4}\qquad (\star)\\ \int_{0}^{+\infty}\frac{\ln x}{(1+x^2)^2}dx&=-\frac{\pi}{4}.\qquad (\star) \end{align} Willing to use only the pole in the upper-half plane, one can proceed as follows: let the angle in the contour you drew approach $\pi$, which gives us $$ \int_{0}^{+\infty}\frac{\ln x}{(1+x^2)^2}dx - \int_{0}^{+\infty}\frac{\ln x +i\pi}{(1+x^2)^2}e^{i\pi}dx=i2\pi \left[i\frac{\pi}{8}\left(\frac{\pi}{2}+2i\right)\right] $$ or $$ 2\int_{0}^{+\infty}\frac{\ln x}{(1+x^2)^2}dx+i\pi \int_0^{+\infty}\frac{1}{(1+x^2)^2}dx=-\frac{\pi}{2}+i\frac{\pi^2}{4}; $$ again, confrontation leads to $(\star)$.

One can avoid computing the annoying double poles and reduce the argument to simple poles at the price of computing a derivative: $$ \int_0^{+\infty}\frac{\ln x}{(1+x^2)^2}dx=-\frac{1}{2}\frac{d}{da}\left(\int_0^{+\infty}\frac{\ln x}{a^2+x^2}dx\right)_{a=1}; $$ now, in the same spirit as above, for an indented half-circle contour in the upper-half plane $$ \int_0^{+\infty}\frac{\ln x}{a^2+x^2}dx-\int_0^{+\infty}\frac{\ln x+i\pi}{a^2+x^2}e^{i\pi}dx=\frac{\pi}{a}\ln a+i\frac{\pi^2}{2a} $$ hence, real and imaginary parts give: $$ \int_0^{+\infty}\frac{\ln x}{x^2+a^2}dx=\frac{\pi}{2a}\ln a\\ \int_0^{+\infty}\frac{dx}{a^2+x^2}=\frac{\pi}{2a}. $$ Performing the above derivative calculated at $a=1$, one recovers the desired results $(\star)$.

  • $\begingroup$ Although the result you derived is certainly correct, it uses a standard keyhole contour (and bravo for using it correctly) rather than the contour specified by the OP. If you had used that contour, you would presumably not need the residue at the Pole $z=e^{i 3 \pi/2}$. $\endgroup$ – Ron Gordon Jan 15 '16 at 22:37
  • $\begingroup$ @RonGordon Thanks; so, you think the OP wanted the contour to be mandatorily enclosed in the sector $[0, 3\pi/4]$? It wasn't clear to me, at first, if the keyhole was acceptable, but then I thought: since $r$ and $R$ will be certainly sent to $0$ and $+\infty$ respectively, there is no reason to keep the angle fixed and not send it to $2\pi$, getting a standard keyhole. $\endgroup$ – Brightsun Jan 15 '16 at 22:44
  • $\begingroup$ To be honest, I have no idea what the OP is after - or, to be more precise, the OP's teacher, as the OP is merely posting a HW question and has yet to demonstrate any independent thought in this forum. But given the drawing, maybe - maybe - the contour is defined so that only one residue is needed, and maybe it can be shown that the angle of the second segment simply doesn't matter. It shouldn't. $\endgroup$ – Ron Gordon Jan 15 '16 at 22:47
  • $\begingroup$ @RonGordon Ok, I see your point. However, I think my argument is made stronger by the fact that the OP('s teacher) did not specify the angle explicitly, whereas he/she took care of setting $r<1<R$. I posted this answer also because the previous ones didn't really take the request of contour integration into account. $\endgroup$ – Brightsun Jan 15 '16 at 22:57
  • $\begingroup$ @RonGordon I added a methond using a contour which is more similar to the one proposed by the OP, in that it only uses the pole at $+i$, as you suggested. It is still angle-dependent, though: let me know what you think! $\endgroup$ – Brightsun Jan 16 '16 at 11:48

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