1
$\begingroup$

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 ?

$\endgroup$
6
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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 $\frac{\Gamma'(z)}{\Gamma(z)}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ah yeah gamma's derivative ! Thank you ! $\endgroup$ – Atmos Dec 28 '17 at 0:53
0
$\begingroup$

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 $$ \int_0^\infty (\ln t)e^{-xt}dt=-\frac{1}{x}(\ln x-\psi^{(0)}(1))=-\frac{1}{x}(\ln x+\gamma). $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.