Can the integral

$$ \int_0^\infty \frac{\log(x)-\log(a)}{x-a}e^{-x} \mathrm{d}x $$

be expressed in terms of some simple special function? I have searched through integral tables but couldn't find anything.

I am mainly interested in values $a \geq 1$, but the integral should be well-defined for all $a>0$.

  • $\begingroup$ Perhaps this is helpful:$$\int^\infty_0\dfrac{\log(x)-\log(a)}{x-a}e^{-x}dx=-\int^\infty_0\dfrac{\log(x)-\log(a)}{x-a}de^{-x}$$ $$=-\lim_{b\to\infty}e^{-x}\dfrac{\log(x)-\log(a)}{x-a}|^b_0+\int^\infty_0e^{-x}d\left(\dfrac{\log(x)-\log(a)}{x-a}\right)$$ Now, $$d\left(\dfrac{\log(x)-\log(a)}{x-a}\right)=\dfrac{(x-a)/x-\log(x)-\log(a)}{(x-a)^2}dx$$ $\endgroup$ – user122283 Mar 26 '14 at 19:57

This integral may be interpreted as a Laplace transform. To see this, let $u=x/a$. Then, $$ \int_0^\infty \frac{\log{x}-\log{a}}{x-a} \mathrm{e}^{-x}\,dx = \int_0^\infty \frac{\log{u}}{u-1} \mathrm{e}^{-au}\,du.$$

In other words, if you knew the Laplace transform of $\frac{\log{x}}{x-1}$, this integral would be solved. Mathematica can do this, but the result is not helpful:

LaplaceTransform[Log[x]/(x - 1), x, a]

$$ e^{-a} \left(-G_{2,3}^{3,0}\left(-a\left| \begin{array}{c} 1,1 \\ 0,0,0 \\ \end{array} \right.\right)+(\log (a)+\gamma +2 i \pi ) (-\Gamma (0,-a))\right) $$

Edit: Summarizing the comments below, the branch cut that Mathematica chose apparently introduced an imaginary component. Let's fix this. First, we re-rewrite the integral as $$ \mathrm{e}^{-a}\left[\int_{-1}^0 + \int_0^\infty\right]\left\{\frac{\log(x+1)}{x}\mathrm{e}^{-ax}\,dx\right\}.$$

The second integral in that can be taken exactly by Mathematica to be

MeijerG[{{0}, {1}}, {{0, 0, 0}, {}}, a]

The first integral, however, let us write that as $I(a)$. Clearly, $$\frac{d^n I}{da^n} = -\int_0^1 x^{n-1}\log(1-x)\mathrm{e}^{ax}\,dx$$

We can (mostly) evaluate these when $a=0$; so let us Taylor expand $I(a)$ about $a=0$. We will obtain $$ I(a) = \frac{\pi^2}{6} + \sum_{n=1}^\infty \frac{H_n}{(n+1)!} a^n $$ where $H_n$ denotes the $n$ Harmonic number. (Partial sums of the harmonic series.)

I don't think we can improve much on this; to summarize, the integral, as originally asked, should be $$ \mathrm{e}^{-a}G_{2,3}^{3,1}\left(a\left| \begin{array}{c} 0,1 \\ 0,0,0 \\ \end{array} \right.\right) + \mathrm{e}^{-a}\left[\frac{\pi^2}{6} + \sum_{n=1}^\infty \frac{H_n}{(n+1)!} a^n\right]. $$

  • $\begingroup$ Thanks, but could there be a glitch in the formula you are reporting? The imaginary parts don't seem to cancel out, yet the integral is clearly real-valued. For $a=1$ for example, WolframAlpha gives me -MeijerG[{{},{1,1}},{{0,0,0},{}},-1] $$ -G_{2,3}^{3,0}\left(-1\left|\begin{array}{c} 1,1 \\ 0,0,0 \\\end{array}\right.\right) = 2.799\dots - i \cdot 1.813\dots $$ whereas (EulerGamma+2*i*Pi)*Gamma[0,-1] $$ (\gamma + 2 i \pi ) \Gamma(0,-1) = 18.645\dots - i \cdot 13.720\dots $$ Could you please double-check? $\endgroup$ – jens Mar 27 '14 at 1:14
  • $\begingroup$ I can only imagine that Mathematica inserted the $2\pi i$ as a choice of branch cut. (I'm not familiar at all with this MeijerG function, so unfortunately cannot give much input.) On the brighter side, if you're interested in the limits of $a\to0$ or $a\to\infty$, there are some very nice asymptotic approximation techniques available to you. $\endgroup$ – Jason Mar 27 '14 at 19:02
  • $\begingroup$ Okay, here is some Mathematica wrangling. First, we rewrite the integral as: $$ \mathrm{e}^{-a} \left[ \int_{-1}^0 \frac{\log(x+1)}{x}\mathrm{e}^{-ax}\,dx + \int_0^\infty \frac{\log(x+1)}{x}\mathrm{e}^{-ax}\,dx \right]. $$ Mathematica takes the 2nd integral in that exactly as >> MeijerG[{{0}, {1}}, {{0, 0, 0}, {}}, a] Then, for $[-1,0]$, we may rewrite that part as: $$ -\int_0^1 \frac{\log(1-x)}{x}\mathrm{e}^{ax}\,dx. $$ At this point, I think you'll have to appeal to an asymptotic argument to estimate this integral. But this at least restores the "realness" that we should have. $\endgroup$ – Jason Mar 27 '14 at 19:17
  • $\begingroup$ Even more cuteness, at $a=0$, we have exactly $\pi^2/6$. (Easily seen by Taylor series.) $\endgroup$ – Jason Mar 27 '14 at 19:34
  • $\begingroup$ Thanks for your answer! I was wondering whether that Meijer G-function could be demystified a little, so I ran WolframAlpha and deduced that $$ G_{2,3}^{3,1}\left(a\left|\begin{array}{c} 0,1 \\ 0,0,0 \\\end{array}\right.\right) = -\dfrac{\gamma^2}{2} + \dfrac{\pi^2}{4} - \gamma\log(a) - \dfrac{\log^2(a)}{2} + \gamma\operatorname{Ei}(a) + \operatorname{Ei}(a)\log(a) - 2a \cdot {}_3F_3(1,1,1;2,2,2;a) - \dfrac{1}{2}\left.\dfrac{\partial^2{}_1F_1(x,y,z)}{\partial x^2}\right|_{(x,y,z)=(0,1,a)} $$ $\endgroup$ – jens Mar 28 '14 at 12:13

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