3
$\begingroup$

$\newcommand{\dx}{\mathrm dx\,}$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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.