This question was part of my complex analysis assignment and I am not able to solve it.
Find the value of $\Gamma (0^+)$ . (It could be $-\infty $ or $+\infty$.)
I used the formula of gamma function which is $\Gamma (z) = \int_{0}^{\infty} t^{z-1}e^{-t} dt $ and I got by putting $z =0^+ $, $\Gamma (0^+)=\int_{0}^{\infty}(1/x ) e^{-x} dx$ and if I integrate it by parts I get
it equal $-\infty-\int_{0}^{\infty}(1/x^2) e^{-x}dx$ .
If I again use integrating by parts to $\int_{0}^{\infty}(1/x^2) e^{-x}dx$ and do it sucessively the power of $x$ will become more negative. So, I am not able to solve the integral.
Can you please tell how to do it?
Best Answer
$$\Gamma(0)=\int_0^\infty \frac{e^{-x}}{x}\mathrm{d}x$$ Let $u=\log(x)$. Then $$\Gamma(0)=\int_{-\infty}^\infty \exp(-e^u)\mathrm{d}u$$ Which certainly diverges to $+\infty$ since the integrand is $\approx 1$ for $u\to-\infty$ and $\approx 0$ for $u\to\infty$.