0
$\begingroup$

For quite sometime I have been stuck. I do not know how to approach this problem. Bear with me if this is already asked:

How can I evaluate $\int_0^\infty\ln(x)e^{-x} dx$?

My trial so far is to use integration by parts but i have no idea how to do it:

$$ \begin{aligned} \text{let } U=&~\ln(x) &dv=e^{-x}dx\\ du =&~\frac{1}{x}dx&V=-e^{-x}\\ \int_0^\infty \ln(x)e^xdx =& -\ln(x)e^{-x}{\LARGE|}_{0}^\infty +\int_0^\infty \frac{e^{-x}}{x}dx \end{aligned} $$

But this doesn't seem right since the first part seems to go to infinity instead of a constant value. Also iI cannot even do the second part. Using R's integrate(function(x) log(x)*exp(-x), 0, Inf) I get -0.57721567 which shows that this integration exists. Can someone help me out?

$\endgroup$
5
  • 3
    $\begingroup$ Your original integral diverges on the $\to\infty$ side, so this can't have an answer. Did you mean $e^{-x}$ ? $\endgroup$
    – Ninad Munshi
    Commented Aug 28, 2020 at 4:14
  • 1
    $\begingroup$ There seems to be a discrepancy. Is it $lnx.e^x$ or $lnx.e^{-x}$ ? $\endgroup$
    – gemspark
    Commented Aug 28, 2020 at 4:19
  • $\begingroup$ @NinadMunshi you are right. It should be $e^{-x}$ $\endgroup$
    – Onyambu
    Commented Aug 28, 2020 at 4:24
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Andrei
    Commented Aug 28, 2020 at 4:26
  • $\begingroup$ The value of your integral is $-\gamma$, where $\gamma$ is the Euler-Mascheroni Constant. It can be taken as the definition of $\gamma$. $\endgroup$
    – Mark Viola
    Commented Aug 28, 2020 at 4:29

1 Answer 1

Reset to default
4
$\begingroup$

By Feynman trick we have that

$$\int_0^\infty\log x \: e^{-x}\:dx = \frac{d}{da}\int_0^\infty x^a e^{-x}\:dx\Biggr|_{a=0} = \Gamma'(1)$$

where $\Gamma(a)$ is the Gamma function.

$\endgroup$
6
  • $\begingroup$ But what is $a$ ? I need to get -0.57721567 as the answer $\endgroup$
    – Onyambu
    Commented Aug 28, 2020 at 4:25
  • $\begingroup$ why did you change the $log(x)$ to $x^a$ ? Are there any links? Sorry I am not that good in math. Also in my trial above, Is there a way to approach the second integral? $\endgroup$
    – Onyambu
    Commented Aug 28, 2020 at 4:36
  • $\begingroup$ Thanks for the trick. Looked through and I do understand now $\endgroup$
    – Onyambu
    Commented Aug 28, 2020 at 4:55
  • $\begingroup$ @Onyambu because $\log x \cdot x^a \implies \frac{d}{da}x^a$ by derivative rules. We just insert an $x^a$ in the original integral since $x^0=1$ was there all along. Your integral above is a transcendental function called exponential integral function. But the integration by parts adds a divergence at $x=0$ where none existed in the original integral. $\endgroup$
    – Ninad Munshi
    Commented Aug 28, 2020 at 4:56
  • $\begingroup$ Thanks so much. If you do not mind, could you share some light with regards to proving the exponential integral function? or do we just take that as a definition? $\endgroup$
    – Onyambu
    Commented Aug 28, 2020 at 5:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .