# Quick question why setting $a=0$ gives an indeterminate solution

$\newcommand{\dx}{\mathrm dx\,}$I’m having lots of trouble figuring this out, so perhaps you guys can help me. For example, let’s take the integral

$$\int\limits_0^{\infty}\dx\frac {\sin x\log x}x=-\frac {\gamma\pi}2$$

Our integral can be computed using differentiation under the integral sign and taking the imaginary part of $$\mathfrak{I}(s)=-\int\limits_0^{\infty}\dx x^{a-1}e^{-sx}=-s^{-a}\Gamma(a)$$ Set $s=i$ and take the imaginary part to get $$\operatorname{Im}\mathfrak{I}(i)=\Gamma(a)\sin\frac {\pi a}2$$ But when I differentiate and set $a=0$, then the gamma function becomes undefined because $\Gamma’(0)$ doesn’t produce a determinate form.

I’m not exactly sure what went wrong. Perhaps you can help me?

From Laplace transform we know that $$\int_0^\infty e^{-sx}x^{a-1}dx=\dfrac{\Gamma(a)}{s^a}$$ is valid where ${\bf Re}(s)>0$, so your reasoning about substitution $s=i$ is false here.