I need help with this integral

$$\int_0^{\infty}e^{-x} \sin x \log x ~ dx$$

WolframAlpha gives the answer to be $\displaystyle \frac{1}{8} (-4 \gamma + π - 2\log(2))$, a curious expression with three fundamental constants.

  • $\begingroup$ Wolfram also gives an expression for the indefinite integral, have you seen that one as well? $\endgroup$ Jun 16, 2018 at 22:03

4 Answers 4



Using Frullani's integral to write $\log(x)=\int_0^\infty \frac{e^{-t}-e^{-xt}}{t}$ reveals

$$\begin{align} \int_0^\infty e^{-x}\sin(x)\log(x)\,dx&=\int_0^\infty e^{-x}\sin(x)\int_0^\infty \frac{e^{-t}-e^{-xt}}{t}\,dt\,dx\\\\ &=\int_0^\infty \frac1t \int_0^\infty \sin(x)(e^{-(x+t)}-e^{-x(t+1)})\,dx\,dt\\\\ &=\int_0^\infty \left(\frac{e^{-t}}{2t}-\frac{1}{t(t^2+2t+2)}\right)\,dt\tag1 \end{align}$$

Next, we see that

$$\int_\epsilon^\infty \frac{e^{-t}}{2t}\,dt=-\frac12e^{-\epsilon}\log(\epsilon)+\frac12\int_\epsilon^\infty e^{-t}\log(t)\,dt\tag2$$


$$\int_\epsilon^\infty \frac{1}{t(t^2+2t+2)}\,dt=-\frac12\log(\epsilon)+\frac14 \log(\epsilon^2+2\epsilon+2)+\frac12\arctan(\epsilon+1)-\frac14\pi\tag3$$

Subtracting $(3)$ from $(2)$ and letting $\epsilon\to 0$, we see that $(1)$ is

$$\int_0^\infty e^{-x}\sin(x)\log(x)\,dx=-\frac\gamma2 -\frac14\log(2)+\frac\pi8$$

as was to be shown!


As an alternative approach we can write

$$\begin{align} \int_0^\infty e^{-x}\sin(x)\log(x)\,dx&=\text{Im}\left(\int_0^\infty e^{-(1-i)x}\log(x)\,dx\right)\\\\ &=\text{Im}\left(\frac1{1-i}\int_0^{(1-i)\infty} e^{-x}\log\left(\frac{x}{1-i}\right)\,dx\right)\\\\ &=\text{Im}\left(\frac1{1-i}\int_0^{(1-i)\infty} e^{-x}\log(x)\,dx-\frac{\log(1-i)}{1-i}\int_0^{(1-i)\infty}e^{-x}\,dx\right)\tag4\\\\ &=\text{Im}\left(\frac1{1-i}\int_0^{\infty} e^{-x}\log(x)\,dx-\frac{\log(1-i)}{1-i}\int_0^{\infty}e^{-x}\,dx\right)\tag5\\\\ &=-\frac\gamma2-\frac14\log(2)+\frac\pi8 \end{align}$$

as expected.

We used Cauchy's Integral Theorem to deform the contour in going from $(4)$ to $(5)$. And we chose to use the principal branch of the logarithm to evaluate $\log(x)$ and $\log(1-i)$.

  • $\begingroup$ What made you think to use Frullani's integral? Indeed a tricky substitution. $\endgroup$
    – Wesley
    Jun 16, 2018 at 22:39
  • $\begingroup$ The presence of $\log(x)$ motivated my using Frullani. $\endgroup$
    – Mark Viola
    Jun 16, 2018 at 22:40

Yet another way, through the Laplace transform.
These preliminary lemmas just follow from integration by parts: $$\mathcal{L}\left(\log x+\gamma\right)(s) = -\frac{\log s}{s},\tag{1} $$ $$\mathcal{L}\left(x e^{-x}\sin x\right)(s) = \frac{2(1+s)}{(2+2s+s^2)^2}\tag{2} $$ and they ensure that $$ \int_{0}^{+\infty}xe^{-x}\sin(x)\cdot\frac{\log x}{x}\,dx = -\int_{0}^{+\infty}(\log s+\gamma)\frac{2(1+s)}{(2+2s+s^2)^2}\,ds\tag{3}$$ where the last integral is straightforward to compute by partial fraction decomposition and integration by parts: $$ \int_{0}^{+\infty}\frac{2(1+s)}{(2+2s+s^2)^2}\,ds = \frac{1}{2},\qquad \int_{0}^{+\infty}\log(s)\frac{2(1+s)}{(2+2s+s^2)^2}\,ds =\frac{2\log 2-\pi}{8}.\tag{4}$$

  • $\begingroup$ For line (3), how did you get from the LHS to the RHS? I have some experience with Laplace transforms, but I'm not quite seeing it. $\endgroup$
    – Wesley
    Jun 18, 2018 at 18:08
  • 1
    $\begingroup$ @MilesDavis: $\mathcal{L}$ is a self-adjoint operator, see en.wikipedia.org/wiki/… $\endgroup$ Jun 18, 2018 at 18:10

$\newcommand{\Log}{\operatorname{Log}}\newcommand{\Im}{\mathfrak{Im}}$Here is an approach using integration under the integral sign (Feynman's trick)

First of all, let $I$ be the integral in the question. One has a term $\log(x)$ but that is also the extra term one gets after differentiating $x^{t}$ with respect to $t$, so that is what inspired me to write this.

Define the function: \begin{align} G(t) := \int^\infty_0 x^{t-1}\sin(x) e^{-x}\,dx \end{align} So by Feynman one gets (verify this): \begin{align} G'(1) = I \end{align}

The problem we have right now is: can we find $G(t)$ in a form we can do something with? The answer is yes. You either see it as Mellin transform, i.e. $G(t)=\mathcal M [e^{-x}\sin(x)] (t)$ and know it by heart (I don't to be honest). Or you see the "$\Gamma$-function" in it. What I mean with the latter is the following, notice that: \begin{align} G(t) = \Im \left(\int^\infty_0 x^{t-1}e^{(i-1)x}\,dx \right) \end{align} Consider the circle sector contour with angle $\pi/4$ on the second quadrant to "translate" that in an expression with $\Gamma$-function (the details are straightforward computations which I leave for you). Using that one can conclude: \begin{align*} \int^\infty_0 x^{t-1}e^{(i-1)x}\,dx= e^{-t\log(2)/2} e^{i\frac \pi 4 t} \Gamma(t) \end{align*} By taking the imaginary part of that one gets: \begin{align} G(t) = e^{-t\log(2)/2}\sin\left( \frac \pi 4 t\right)\Gamma(t) \end{align} Hence: \begin{align} G'(t) = -\frac{\log(2)}{2} e^{-t\log(2)/2}\sin\left( \frac \pi 4 t\right)\Gamma(t) + \frac{\pi}{4}e^{-t\log(2)/2}\cos\left( \frac \pi 4 t\right)\Gamma(t) + e^{-t\log(2)/2}\sin\left( \frac \pi 4 t\right)\Gamma'(t) \end{align} Now we almost have the result, namely: \begin{align*} I = G'(1) &= -\frac{\log(2)}{2} e^{-\log(2)/2}\frac{\sqrt[]{2}}{2} + \frac{\pi}{4}e^{-\log(2)/2}\frac{\sqrt[]{2}}{2} + e^{-\log(2)/2}\frac{\sqrt[]{2}}{2}\Gamma'(1)\\ &= \frac{1}{2}\left[ -\frac{\log(2)}{2} + \frac{\pi}{4}+\Gamma'(1) \right] \end{align*} By using Feynman's trick (again): \begin{align*} \Gamma'(1) = \int^\infty_0 \log(x)e^{-x}\,dx = -\gamma \end{align*} Finally we can conclude that: \begin{align*} \int^\infty_0 e^{-x}\sin (x)\log (x)\,dx = \frac{1}{8}\left( -2\log(2) + \pi-4\gamma \right) \end{align*}


Partial solution. Use $\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$ and switch the sum and integral (I can't justify this yet) to get $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \int_0^\infty e^{-x}x^{2n+1}\log(x)dx.$$ If we let $I_m = \int_0^\infty e^{-x}x^{m}\log(x)dx$, then a simple integration by parts yields $I_m = mI_{m-1}+(m-1)!$ which, together with the fact that $I_0 = -\gamma$, gives $I_m = -m!\gamma+a_m$ where the sequence $(a_m)_m$ satisfies $a_m = ma_{m-1}+(m-1)!$ [these are known as the unsigned stirling numbers of the first kind]. So, we just need to evaluate $$\sum_{n=0}^\infty (-1)^n\left[-\gamma+\frac{a_{2n+1}}{(2n+1)!}\right].$$ Now maybe there's some known fact about these unsigned stirling numbers that finishes the job.


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.