Prove that $\int_0^{\infty} \frac{\ln x \,dx}{x^b(x+1)}=\pi ^2 \cot(\pi b) \csc(\pi b)$

Question

Prove that $$\int_0^{\infty} \frac{\ln x}{x^b(x+1)} dx=\pi ^2 \cot(\pi b) \csc(\pi b)$$ for $0<b<1$. It is not hard to show it equals to 0 when $b=1/2$, but I don't know what to do next.

Answer 1

Let $\dfrac{1}{1+x}=u$ then \begin{align} \int_0^{\infty} \frac{\ln x}{x^b(x+1)} dx &= -\dfrac{d}{db}\int_0^{\infty} \frac{1}{x^b(x+1)} dx \\ &= -\dfrac{d}{db}\int_0^1 u^{b-1}(1-u)^{-b} du \\ &= -\dfrac{d}{db}\left(\Gamma(b)\Gamma(1-b)\right) \\ &= -\dfrac{d}{db}\left(\dfrac{\pi}{\sin\pi b}\right) \\ &= \color{blue}{\pi ^2 \cot\pi b\csc\pi b} \end{align}

Answer 2

We integrate with $b$ a real from $(0,1)$

$$f(z) = \frac{\mathrm{Log}(z)}{z+1} \exp(-b\times \mathrm{Log}(z))$$

around a keyhole contour with the slot on the positive real axis, which is also where the branch cut of the logarithm is located (argument of the logarithm is between $0$ and $2\pi$). Now for the large circle we get $\lim_{R\to\infty} 2\pi R \log R / R^b /R = 0$ so there is no contribution in the limit. For the small circle around the origin we find $\lim_{\epsilon\to\ 0} 2\pi \epsilon \log \epsilon / \epsilon^b = \lim_{\epsilon\to\ 0} 2\pi \epsilon^{1-b} \log \epsilon = 0$ so there is no contribution here either.

We get for the upper line segment

$$\int_0^\infty \frac{\log x}{x+1} \exp(-b\times \log x) \; dx$$

which is our target integral, call it $J$. The lower line segment contributes

$$-\int_0^\infty \frac{\log x + 2\pi i}{x+1} \exp(-b\times \log x) \exp(-b\times 2\pi i) \; dx \\ = - \exp(-b\times 2\pi i) \int_0^\infty \frac{\log x + 2\pi i}{x+1} \exp(-b\times \log x) \; dx \\ = - \exp(-b\times 2\pi i) J - \exp(-b\times 2\pi i) 2\pi i \int_0^\infty \frac{1}{x+1} \exp(-b\times \log x) \; dx \\ = - \exp(-b\times 2\pi i) J - \exp(-b\times 2\pi i) 2\pi i K.$$

where $J$ and $K$ are real numbers. We thus have

$$(1- \exp(-b\times 2\pi i))J - \exp(-b\times 2\pi i) 2\pi i K = 2\pi i \mathrm{Res}_{z=-1} f(z).$$

The residue now yields

$$(1- \exp(-b\times 2\pi i))J - \exp(-b\times 2\pi i) 2\pi i K = 2\pi i \times \pi i \exp(-b \times \pi i)$$

or

$$(\exp(b\times \pi i) - \exp(-b\times \pi i))J - \exp(-b\times \pi i) 2\pi i K = 2\pi i \times \pi i$$

and finally

$$\sin(\pi b) J - \exp(-b\times \pi i) \pi K = \pi \times \pi i.$$

Using the same contour and the same branch of the logarithm to compute $K$ we find with

$$g(z) = \frac{1}{z+1} \exp(-b\times \mathrm{Log}(z))$$

that

$$(1-\exp(-b \times 2\pi i)) K = 2\pi i \times \mathrm{Res}_{z=-1} g(z) = 2\pi i \times \exp(-b \times \pi i)$$

which yields

$$2i \sin(\pi b) K = 2\pi i \quad\text{so that}\quad K = \frac{\pi}{\sin(\pi b)}.$$

Taking the real part of the equation linking $J$ and $K$ we get

$$\sin(\pi b) J - \cos(-\pi b) \frac{\pi^2}{\sin(\pi b)} = 0$$

or

$$\bbox[5px,border:2px solid #00A000]{ J = \pi^2 \cot(\pi b) \csc(\pi b)}$$

as claimed.

Answer 3

You might start with $$\int_0^\infty \dfrac{dx}{x^b(x+1)} = \pi \csc(\pi b)$$ and differentiate with respect to $b$.