Prove that $\int_0^\infty e^{-x} \ln x d x = - \gamma $

Question

I can see it is right by using some knowledge of the Gamma function. We have

$$ \Gamma(\alpha ) = \int_0^\infty e^{-x} x^{\alpha - 1 } dx . $$

Differentiating with respect to $\alpha$, we get

$$ \frac{d\Gamma}{d \alpha} = \int_0^\infty e^{-x} x^{\alpha - 1 }\ln x dx . $$

Setting $\alpha = 1 $, we get

$$ \int_0^\infty e^{-x} \ln x dx = \frac{d\Gamma}{d\alpha }\bigg|_{\alpha = 1 }= - \gamma. $$

But the knowledge $d \Gamma/d \alpha |_{\alpha = 1}=-\gamma$ is a mystery to me.

Can anyone find an elementary proof?

Answer 1

You can prove that $$\lim I_n=\int_0^\infty e^{-x}\ln x\,dx$$ using the dominated convergence theorem, where $$I_n=\int_0^n\left(1-\frac xn\right)^n\ln x\,dx.$$ Now substitute $y=1-x/n$. Then \begin{align} I_n&=n\int_0^1y^n\ln(n(1-y))\,dy\\ &=n\ln n\int_0^1 y^n\,dy+n\int_0^1y^n\ln(1-y)\,dy\\ &=\frac n{n+1}\left(\ln n-\int_0^1\frac{1-y^{n+1}}{1-y}\,dy\right)\\ &=\frac n{n+1}\left(\ln n-\sum_{k=1}^{n+1}\frac1k\right) \end{align} where we have integrated by parts along the way. This tends to $-\gamma$.

Answer 2

Here we will address your integral: \begin{equation} I = \int_0^\infty e^{-x} \ln(x)\:dx \nonumber \end{equation}

To do so we use the fact that: \begin{equation} \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a = \ln(x) \end{equation}

As such (and by Leibniz's Integral Rule): \begin{align} I &= \int_0^\infty e^{-x} \ln(x)\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a}\int_0^\infty e^{-x} x^a\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \Gamma(a + 1) \nonumber \\ &= \lim_{a \rightarrow 0^+} \Gamma'(a + 1) = \lim_{a \rightarrow 0^+} \Gamma(a + 1)\:\psi^{(0)}(a + 1)\nonumber \\ &= \Gamma(1)\:\psi^{(0)}(0) = \psi^{(0)}(0) = -\gamma \end{align}

Answer 3

Your method seems like a reasonable way to derive this result, and you are almost there. Now use the Taylor expansion of the digamma function, which is:

$$\psi(z+1) = -\gamma - \sum_{k=1}^\infty \zeta(k+1) \cdot (-z)^k.$$

Substituting $z=0$ gives:

$$\psi(1) = -\gamma - \sum_{k=1}^\infty \zeta(k+1) \cdot 0^k = -\gamma.$$

So you have:

$$\int_0^\infty e^{-x} \ln x dx = \Gamma'(1) = \psi(1) \cdot \Gamma(1) = -\gamma \cdot 1 = -\gamma.$$