differential Gamma function

Question

The Gamma function is $\Gamma(t)=\int^\infty_0x^{t-1}e^{-x}dx$.

In some book, it writes that $\Gamma'(1)=\int^1_0(e^{-x}-1)dx/x+\int^\infty_1e^{-x}dx/x$.

Q: How to show that ?

Answer 1

An alternative derivation. We have: $$ \psi(x) = \frac{d}{dx}\log\Gamma(x) = \frac{\Gamma'(x)}{\Gamma(x)}\tag{1} $$ but from the Weierstrass and Euler products for the $\Gamma$ function we have: $$ \gamma=\sum_{n\geq 1}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right),\qquad \psi(x)=-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+x-1}\right)\tag{2}$$ so that: $$ \Gamma'(1) = -\gamma = \sum_{n\geq 1}\left(\log\left(1+\frac{1}{n}\right)-\frac{1}{n}\right)\tag{3} $$ Now we may exploit $\frac{1}{n}=\int_{0}^{+\infty}e^{-nx}\,dx$ and Frullani's theorem $\log\left(1+\frac{1}{n}\right)=\int_{0}^{+\infty}\frac{1-e^{-x}}{x}e^{-nx}\,dx$ to give an integral representation to the LHS of $(3)$: $$ \Gamma'(1) = \int_{0}^{+\infty}\left(\frac{1-e^{-x}}{x}-1\right)\sum_{n\geq 1}e^{-nx}\,dx = \int_{0}^{+\infty}\left(\frac{1}{xe^x}-\frac{1}{e^x-1}\right)\,dx\tag{4}$$ as wanted. Your integral representation is actually a bit different, $$ \Gamma'(1)=\int_{0}^{+\infty}\frac{e^{-x}-\mathbb{1}_{x\leq 1}(x)}{x}\,dx \tag{5}$$ but can be converted into $(4)$ by considering the Laplace transform of $e^{-x}-\mathbb{1}_{x\leq 1}(x)$ and the inverse Laplace transform of $\frac{1}{x}$. An interesting consequence of $(5)$ is

$$ \gamma = \sum_{n\geq 0}\frac{(-1)^n}{(n+1)(n+1)!}-\int_{1}^{+\infty}\frac{dx}{x e^x} \tag{6}$$ allowing us to approximate $\gamma$ with decent accuracy.

Answer 2

HINT, when $\Re\left(t\right)>0$:

$$\Gamma\space'\left(1\right):=\lim_{t\to1}\space\frac{\text{d}}{\text{d}t}\left\{\int_0^\infty\frac{x^{t-1}}{e^x}\space\text{d}x\right\}=\int_0^\infty\frac{\ln\left(x\right)x^{1-1}}{e^x}\space\text{d}x=\int_0^\infty\frac{\ln\left(x\right)}{e^x}\space\text{d}x=-\gamma\tag1$$