Show that for \( b>1 \) we have
$$ \int^\infty_0 \frac{\ln(x)}{x^b-1}\, dx = \frac{\pi^2}{b^2} \csc^2\left( \frac{\pi}{b}\right)$$
Proof
Consider the following function
$$f(z) = \frac{\log(z)}{e^{b\log(z)}-1}$$
Now consider a sector with an angle of \(\frac{\pi}{b}\)
The integration of the whole sector is \(0 \) since the function is analytic in and on the contour by taking the principle branch of the logarithm .
The integration along the x-axis is
$$\int^{R}_{r} \frac{\log(x)}{x^b-1}$$
The integration along the big and small circular sectors
It can be proven easily by taking \( r\to 0 , R \to \infty\) that they vanish.
The integration along the tilted line
Can be parametrized using the transformation \(z=te^{i\frac{\pi}{b}}\) , \( R<t<r\)
$$-e^{\frac{\pi i}{b}}\int^{R}_r \frac{\log(te^{\frac{\pi i}{b}})}{e^{b\log\left(\frac{\pi i}{b} \right)}-1}\, dt$$
$$e^{\frac{\pi i}{b}}\int^{R}_r \frac{\log(t)+i\frac{\pi}{b}}{t^b+1}\, dt$$
$$e^{\frac{\pi i}{b}}\int^{R}_r \frac{\log(t)}{t^b+1}\, dt+ie^{\frac{\pi i}{b}}\int^{R}_0\frac{\frac{\pi}{b}}{t^b+1}\, dt\,\,\, (1)$$
The integrals can be easily found using the beta function by taking the principle value and
\(r\to 0 \,\,\,\, , \,\,\,\, R \to \infty\)
$$e^{\frac{\pi i}{b}} \left( -\frac{\pi^2}{b^2} \csc\left( \frac{\pi}{b}\right) \cot\left( \frac{\pi}{b}\right)+i\frac{\pi^2}{b^2} \csc\left( \frac{\pi}{b}\right)\right)$$
This can be written as
$$-\frac{\pi^2e^{\frac{\pi i}{b}}}{b^2} \csc\left( \frac{\pi}{b}\right) \left( \cot\left( \frac{\pi}{b}\right)-i\right)$$
$$-\frac{\pi^2e^{\frac{\pi i}{b}}}{b^2} \csc^2\left( \frac{\pi}{b}\right) \left( \cos\left( \frac{\pi}{b}\right)-i\sin\left( \frac{\pi}{b}\right)\right)$$
Now using the Euler formula we have
$$-\frac{\pi^2}{b^2} \csc^2\left( \frac{\pi}{b}\right)$$
Summing the curves together and using Cauchy integral formula gives the desired result .
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