Prove that $\int_{0}^{\infty} e^{-x}x\log{x}\ dx =1-\gamma$

Question

I tried to this equation:

$$\int_{0}^{\infty} e^{-x}x\log{x}\ dx =1-\gamma$$

but I was stuck because I couldn't avoid the appearance of $Ei(0)$ and $1/0$ for my bad way.

I want to know how to avoid them and prove this.

Answer 1

Since $\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}dx$, your integral is $\Gamma^\prime(2)=\psi(2)=1-\gamma$, with $\psi$ the digamma function.

Answer 2

It is a known result that ${\int_{0}^{\infty}e^{-x}\log(x)dx=-\gamma}$ (see https://www.youtube.com/watch?v=80_fNEflUdw). We are going to try and exploit this.

Notice ${\frac{d}{dx}\left(e^{-x}\log(x)\right)=\frac{e^{-x}(1-x\log(x))}{x}}$. So what? Well, we can try and use integration by parts along with some trickery. You see that

$${\int_{0}^{\infty}x\frac{d}{dx}\left(e^{-x}\log(x)\right)dx=-\int_{0}^{\infty}e^{-x}\log(x)dx}$$

from integration by parts on the one hand. On the other hand, explicitly using the derivative of ${e^{-x}\log(x)}$ yields

$${\int_{0}^{\infty}x\frac{d}{dx}\left(e^{-x}\log(x)\right)dx= \int_{0}^{\infty}e^{-x} - xe^{-x}\log(x)dx}$$ $${=1-\int_{0}^{\infty}xe^{-x}\log(x)dx}$$

So overall,

$${1-\int_{0}^{\infty}xe^{-x}\log(x)dx=-\int_{0}^{\infty}e^{-x}\log(x)dx = \gamma}$$

Which implies that

$${\int_{0}^{\infty}xe^{-x}\log(x)dx=1-\gamma}$$

Answer 3

Just for fun:

$$\begin{eqnarray*} \int_{0}^{n}\left(1-\frac{x}{n}\right)^n x\log(x)\,dx &=& n^2\int_{0}^{1}(1-x)^n x\log(nx)\,dx\\&=&n^2\log(n)B(n+1,2)+n^2\left.\frac{d}{db}B(n+1,b)\right|_{b=2}\end{eqnarray*}$$ where $$ B(n+1,2)=\frac{\Gamma(n+1)\Gamma(2)}{\Gamma(n+3)}=\frac{1}{(n+1)(n+2)} $$ $$\begin{eqnarray*} \frac{d}{db}B(n+1,b)=\frac{d}{db}\left(\frac{\Gamma(n+1)\Gamma(b)}{\Gamma(n+1+b)}\right)&=&\frac{\Gamma(n+1)\Gamma'(b)}{\Gamma(n+1+b)}-\frac{\Gamma(n+1)\Gamma(b)\Gamma'(n+1+b)}{\Gamma(n+1+b)^2}\\&=&\frac{\Gamma(n+1)\Gamma(b)}{\Gamma(n+1+b)}\left(\psi(b)-\psi(n+1+b)\right)\end{eqnarray*}$$ $$ \left.\frac{d}{db} B(n+1,b)\right|_{b=2} = \frac{\psi(2)-\psi(n+3)}{(n+1)(n+2)} $$ hence by the dominated convergence theorem $$ \int_{0}^{+\infty}e^{-x} x \log(x)\,dx = \lim_{n\to +\infty}\left(\log(n)+\psi(2)-\psi(n+3)\right)=\psi(2)=1-\gamma$$ since $\psi(n)=\log(n)+o(1)$ by Stirling's inequality.