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The task is, considering the following function and contour, to proove that this is the value of the integral. (see the picture)

Click here to see the problem

I managed to show that the integral along two circles vanishes, and that the residue of the pole $z=0$ is equal to $1$.

I ended up with the following expression:

$\int_{1}^{\infty} \frac{log^2(x-1)}{x^3} dx + \int_{\infty}^{1} \frac{log^2((x-1)e^{i2\pi})}{x^3} dx = 2\pi i$

Where $e^{2\pi i}$ is due to the orientation of the branch cut that goes from $1$ to $\infty$.

Now when I expand this I get:

$-4\pi i \int_{0}^{\infty} \frac{log(x-1)}{x^3} dx+4\pi^2\int_{0}^{\infty} \frac{1}{x^3} dx=2\pi i$

Obviously I get the right answer by matching the imaginary part, however I am left with non-zero real part...

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  • $\begingroup$ That problem needs to be posted here. The link will disappear in time. Uploaded images are acceeptabble here. $\endgroup$ – zhw. Jun 11 '18 at 20:55
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Little note: such integral can be computed through a simple symmetry trick. $$\int_{1}^{+\infty}\frac{\log^2(x-1)}{x^3}\,x\stackrel{x\mapsto x^{-1}}{=}\int_{0}^{1}x\left[\log^2(1-x)-2\log(x)\log(1-x)+\log^2(x)\right]\,dx \tag{1}$$ due to $\int_{0}^{1}f(x)\,dx = \frac{1}{2}\int_{0}^{1}f(x)+f(1-x)\,dx$, equals $$ \int_{0}^{1}\log^2(x)\,dx - \int_{0}^{1}\log(x)\log(1-x)\,dx = 2- \int_{0}^{1}\log(x)\log(1-x)\,dx. \tag{2}$$ Since $$ \int_{0}^{1}x^\alpha (1-x)^\beta\,dx = \frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} \tag{3} $$ for any $\alpha,\beta>-1$ due to Euler's Beta function, by applying $\left.\frac{\partial^2}{\partial\alpha\,\partial\beta}(\ldots)\right|_{(\alpha,\beta)=(0,0)}$ to both sides of $(3)$ we get $\int_{0}^{1}\log(x)\log(1-x)\,dx=2-\frac{\pi^2}{6}$ and the original integral simply equals $\color{red}{\zeta(2)}$. The last identity also follows from the functional relations for the dilogarithm. As an elementary alternative,

$$\begin{eqnarray*} \int_{0}^{1}\log(x)\log(1-x)\,dx &=& -\sum_{n\geq 1}\int_{0}^{1}\frac{x^n}{n}\log(x)\,dx=\sum_{n\geq 1}\frac{1}{n(n+1)^2}\\&=&\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^2}\right)\\&=&1-(\zeta(2)-1)=2-\zeta(2).\tag{4}\end{eqnarray*}$$ In particular the computation of $\int_{1}^{+\infty}\frac{\log^2(x-1)}{x^3}\,dx$ through Complex Analysis provides a proof of the Basel problem.

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Okay, I think I have finally found a mistake...

The thing is when calculating residue one should also include factor that comes from the branch point, so when looking at the residue at $z=0$ one should expand:

$\frac{log^2((1-z)e^{i\pi})}{x^3}$

rather than:

$\frac{log^2(1-z)}{x^3}$

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