I would like to solve integrals of the form $$I(c) := \int_0^\infty \log(1+x) x^{-c} \, dx ,$$ where $c \in (1,2)$.
Mathematica says either
1) $I(c) = \frac{\pi}{1-c} \csc(\pi c)$
or
2) $I(c) = \frac{\Gamma(2 - c) \Gamma(c)}{(-1 + c)^2}$
depending on the way I formulate the integral. A plot suggests, that 1) and 2) are equivalent.
Can someone explain how one of these solutions is derived and how 1) and 2) are related?
Thanks a lot!
On
Euler's reflection formula is $\Gamma(z)\Gamma(1-z)=\pi\csc(\pi z).$ We also know that $\Gamma(1+z)=z\Gamma(z)$. Thus, $\Gamma(2-z)=(1-z)\Gamma(1-z)$. Thus, $$\frac{\Gamma(2-z)\Gamma(z)}{1-z}=\pi\csc(\pi z).$$ Then we just divide both sides by $1-z$ to get,$$\frac{\Gamma(2-z)\Gamma(z)}{(1-z)^2}=\frac{\pi}{1-z}\csc(\pi z).$$
First consider the integral \begin{align} B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \int_{0}^{\infty} \frac{t^{x-1} \, dt}{(1+t)^{x+y}} \end{align} which defines the Beta function. Take the derivative with respect to $y$ by using \begin{align} \partial_{y} \frac{1}{(1 + t)^{y}} = \partial_{y} e^{y \, \ln(1+t)} = \frac{\ln(1+t)}{(1+t)^{y}} \end{align} to obtain \begin{align} B(x,y) \, \left( \psi(y) - \psi(x+y) \right) = \int_{0}^{\infty} \frac{t^{x-1} \, \ln(1+t)}{(1+t)^{x+y}} \, dt \end{align} where $\psi(x)$ is the digamma function. Now let $x = 1-c$ to obtain \begin{align} B(1-c,y) \, \left( \psi(y) - \psi(y+1-c) \right) = \int_{0}^{\infty} \frac{t^{-c} \, \ln(1+t)}{(1+t)^{y+1-c}} \, dt. \end{align} Taking the limit as $y \to c-1$ yields \begin{align} \int_{0}^{\infty} t^{-c} \, \ln(1+t) \, dt = \frac{\Gamma(2-c) \, \Gamma(c)}{(1-c)^{2}}. \end{align} Using $\Gamma(x+1) = x \, \Gamma(x)$ then this becomes \begin{align} \int_{0}^{\infty} t^{-c} \, \ln(1+t) \, dt = \frac{\Gamma(1-c) \, \Gamma(c)}{1-c}. \end{align} Using the property \begin{align} \Gamma(1-x) \, \Gamma(x) = \pi \, \csc(\pi x) \end{align} then \begin{align} \int_{0}^{\infty} t^{-c} \, \ln(1+t) \, dt = \frac{\pi \, \csc(\pi \, c)}{1-c}. \end{align}