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I am trying to find

$$\int_{0}^{\infty} {\frac{\ln(x)}{(x+1)^3}}dx$$

Using the residue formula.

I am mainly having a hard time finding a contour that works, since we must include the pole of order three at $x=-1$.

I have tried a partial, indented circle that goes from $\theta = 0$ to $4\pi/3$ and a $3/4$ circle as well, but in both cases, if $\gamma_3(t) = te^{i\theta}, t \in (R,0)$ for a fixed $\theta$ in the third quadrant, we are left with the real part of the integral, as well as a complex integral that is just as difficult to solve.

The residue I calculated is:

$$res_{-1}f(z) = \lim_{z \rightarrow -1}\frac{1}{3!} \frac{d^2}{dz^2} (z+1)^3 \frac{\ln{z}}{(1+z)^3}$$ $$=\frac{1}{2}* \frac{-1}{-1^2} = \frac{-1}{2}$$

So that means the entire complex integral should be $2\pi i* \frac{-1}{2} = -\pi i$...

I had another idea for a contour that catches the pole: we have

$$\gamma_1(t) = t, t \in (\epsilon, R)$$ $$\gamma_2(t) = Re^{it}, t\in (0, \pi/2)$$ $$\gamma_3(t) = e^{it} -1, t \in (\pi, 2\pi - \epsilon)$$ $$\gamma_4(t) = $$ the little tiny circle to complete the contour.

But this one, for $\gamma_3$ gives us a weird integral:

$$\int_{\pi}^{2\pi - \epsilon}\frac{\ln{e^{it} - 1}}{(e^{it} - 1 +1)^3}(ie^{it})$$

Which looks a little better, but I'm not sure how to hande the log function in this case, and it doesn't look promising. Any help would be appreciated.

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The contour we need to use for this question is a basic keyhole contour

Keyhole

First, parametrize about the contour in four different areas. Let the smaller circle have a radius of $\epsilon$ and the larger circle have a radius of $R$. Furthermore, denote the arcs of the larger circle and smaller circle as $\Gamma_R$ and $\gamma_{\epsilon}$ respectively. Considering the function

$$f(z)=\frac {1}{(z+1)^3}$$

We have

$$\begin{multline}\oint\limits_{\mathrm C}dz\, f(z)\log^2z=\int\limits_{\epsilon}^{R}dx\, f(x)\log^2x+\int\limits_{\Gamma_{R}}dz\, f(z)\log^2z\\-\int\limits_{\epsilon}^{R}dx\, f(x)(\log|x|+2\pi i)^2+\int\limits_{\gamma_{\epsilon}}dz\, f(z)\log^2z\end{multline}$$

As $R\to\infty$ and $\epsilon\to0$, the integrals around the arc vanish leaving us with

$$\oint\limits_{\mathrm C}dz\, f(z)\log^2z=-4\pi i\int\limits_0^{\infty}dx\, f(x)\log x+4\pi^2\int\limits_0^{\infty}dx\, f(x)$$

All you have to do left is calculate the residue, multiply that by $2\pi i$, and divide the imaginary part by $-4\pi$ to get the answer to your integral.

I'll leave the rest of the work up to you.

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  • $\begingroup$ Wait, won't the two sides of the "tube" part always cancel out? Also, is this contour valid for the log function? I thought the log can't be defined on a full circle?? $\endgroup$ – pictorexcrucia May 6 '18 at 17:45
  • $\begingroup$ @pictorexcrucia Yes, both sides of the "tube" part will cancel out. So a nice tip to know when dealing with logs is that if your function $f(z)$ has a natural log, it's always nice to consider the function $g(z)=f(z)\log z$. $\endgroup$ – Frank W. May 6 '18 at 17:48
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    $\begingroup$ @pictorexcrucia Yes, this contour is valid for a logarithmic function. And no, this contour is not a full circle. It has that distinct cut along the real axis which is why we call it a keyhole contour. $\endgroup$ – Frank W. May 6 '18 at 17:50
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    $\begingroup$ Actually, the integrals along opposite sides of the branch cut do NOT cancel. We have $\log^2(x)-(\log(x)+2\pi i)^2=-4\pi i \log(x)+4\pi^2\ne 0$. And inside and on the contour, $\log(z)$ is analytic. So, the contour, equipped with the branch cut, is not a closed circle. $\endgroup$ – Mark Viola May 6 '18 at 17:51
  • $\begingroup$ @MarkViola Sorry, but I was assuming he meant the extra power of the log which is $\log^2x$ in this case. Perhaps I shouldn't have said that they "cancel out" $\endgroup$ – Frank W. May 6 '18 at 17:53
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HINT:

Analyze the contour integral $\oint_C \frac{\log^2(z)}{(z+1)^3}\,dz$ where $C$ is the classical keyhole contour with a branch cut along the non-negative real axis.

Use the residue theorem to evaluate this closed-contour integral (there is a third order pole at $z=-1$).

Finally, note that $\log^2(z)=\log^2(x)$ on that part of the branch cut in Quadrant I and $\log^2(z)=(\log(x)+2\pi i)^2$ on that part of the branch cut in Quadrant IV.

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