# Proof that $\int_{0}^{+\infty}\ln\left(t\right)e^{-xt}\text{d}t=-\frac{\ln\left(x\right)+\gamma}{x}$

I wonder how we could prove that $$\int_{0}^{+\infty}\ln\left(t\right)e^{-xt}\text{d}t=-\frac{\ln\left(x\right)+\gamma}{x}$$ Let $f$ the function defined on $\left]0,+\infty\right[$ with $x \in \left]0,+\infty\right[$ by $$f\left(t\right)=\ln\left(t\right)e^{-xt}$$ Hence $$\left|f\left(t\right)\right| \underset{(0^{+})}{\sim}-\ln\left(t\right) \text{ }\text{ and }\text{ }\left|f\left(t\right)\right|\underset{(+\infty)}{=}o\left(\frac{1}{t^2}\right)$$ The first result shows integrability around $0^{+}$ and the second one integrability approaching $+\infty$.\ It assures the existence but how can we determine this value ? I did not try every possibility, I mean using change of variable that could help ... I would like to avoid using complex analysis.

Any ideas ?

Upon applying the substitution $xt\mapsto t$, we have

$$\int_{0}^{\infty} e^{-xt} \log t \, dt = \frac{1}{x}\int_{0}^{\infty} e^{-t} (\log t - \log x) \, dt = \frac{1}{x}\left( -\log x + \int_{0}^{\infty} e^{-t} \log t \, dt \right)$$

and hence it suffices to prove the result when $x=1$. Now notice that

$$\left(1-\frac{t}{n}\right)_+^{n-1} \xrightarrow[n\to\infty]{} e^{-t}$$

where $a_+=\max\{a,0\}$ and it is not hard to check that this sequence is dominated by $Ce^{-t}$ for some absolute constant $C>0$. So by the dominated convergence theorem,

\begin{align*} \int_{0}^{\infty}e^{-t}\log t\,dt &= \lim_{n\to\infty}\int_{0}^{n} \left(1-\frac{t}{n}\right)^{n-1} \log t \, dt \\ {\small (t=n(1-u))} \quad &= \lim_{n\to\infty}\int_{0}^{1} nu^{n-1}(\log n + \log(1-u)) \, du \end{align*}

and we can check that the last integral can be evaluated exactly as

\begin{align*} &\int_{0}^{1} nu^{n-1}(\log n + \log(1-u)) \, du \\ &\hspace{1em} = \log n - \int_{0}^{1} nu^{n-1}\left( \sum_{k=1}^{\infty} \frac{u^k}{k} \right) \, du = \log n - \sum_{k=1}^{\infty} \frac{n}{k} \int_{0}^{1} u^{n+k-1} \, du \\ &\hspace{2em} =\log n - \sum_{k=1}^{\infty} \frac{n}{k(n+k)} =\log n - \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{n+k}\right) \\ &\hspace{3em} =\log n - \sum_{k=1}^{n} \frac{1}{k}. \end{align*}

This converges to $-\gamma$ by the definition of the Euler-Mascheroni constant.

With the change of variable $xt=s$ we get

\begin{align}\int_0^\infty (\ln t)e^{-xt}\,\mathrm dt&=\int_0^\infty (\ln s/x)x^{-1}e^{-s}\,\mathrm ds\\&=\frac1x\left(\int_0^\infty(\ln s)s^0e^{-s}\,\mathrm ds-(\ln x)\int_0^\infty s^0e^{-s}\,\mathrm ds\right)\\&=\frac1x(\Gamma'(1)-(\ln x)\Gamma(1))\end{align}

To evaluate $\Gamma'(1)$ you can use the identity

$$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}k\right)e^{-z/k},\quad\text{for }z\in\Bbb C\setminus(-\Bbb N)$$

and after applying logarithm to the reciprocal and differentiating you find an expression for $\frac{\Gamma'(z)}{\Gamma(z)}$.

• Ah yeah gamma's derivative ! Thank you ! – Atmos Dec 28 '17 at 0:53

By using $$\Gamma(s+1)=\int_0^\infty t^se^{-st}dt$$ one has $$\int_0^\infty t^se^{-xt}dt=x^{-s-1}\Gamma(s+1).$$ So $$\int_0^\infty t^s(\ln t)e^{-xt}dt=-x^{-s-1}(\ln x-\psi^{(0)}(s+1))$$ and hence $$\int_0^\infty (\ln t)e^{-xt}dt=-\frac{1}{x}(\ln x-\psi^{(0)}(1))=-\frac{1}{x}(\ln x+\gamma).$$