Mellin transform integral

Evaluate the integral:

$\mathcal{J} = \int_0^\infty x \log x e^{-\sqrt{x}} \, \mathrm{d}x$

Solution

We are evaluating the Mellin transform of the function $f(x)=e^{-\sqrt{x}}$ .

$\begin{align*} \mathcal{M}\left ( f \right ) &= \int_{0}^{\infty} x^{s-1} e^{-\sqrt{x}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\sqrt{x}}{=\! =\! =\! =\! =\!} 2\int_{0}^{\infty} u^{2s-1} e^{-u} \, \mathrm{d}u\\ &= 2 \Gamma\left ( 2s \right ) \end{align*}$

where $\Gamma$ is the Euler’s Gamma function. Hence,

$\begin{align*} \int_{0}^{\infty} x \log x e^{-\sqrt{x}} \, \mathrm{d}x &= \mathcal{M}'(f) \bigg|_{s=2} \\ &= \left ( 2 \Gamma(2s) \right )'\bigg|_{s=2}\\ &= 4 \Gamma(2s) \psi^{(0)}(2s) \bigg|_{s=2}\\ &=4 \Gamma(4) \psi^{(0)}(4) \\ &= 4 \cdot 6 \cdot \left ( \frac{11}{6} - \gamma \right ) \\ &=44 - 24 \gamma \end{align*}$