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I have this integral

$$\int_0^\infty \frac{x^{2 m}\;\ln^n (x) }{e^{\frac{2 p +1}{2}x}} dx\;\; m,n,p \in \mathbb{N}$$

and I'like to know the general solution.

From WolframAlpha I got some solutions, e.g.

$$\int_0^\infty \frac{x^2\ln(x)}{e^{\frac{3x}{2}}} dx = \frac{8}{27}\left(3-2\gamma-\ln\left(\frac{9}{4}\right)\right) $$ and

$$\int_0^\infty \frac{x^{2}\ln^{2}(x)}{e^{\frac{3x}{2}}} dx = \frac{8}{81}\left(6-18\gamma+6\gamma^{2}+\pi^{2}-6 \ln \left(\frac{3}{2}\right)\left(3-2\gamma-\ln\left(\frac{3}{2}\right)\right) \right) $$

etc... but I'd like to know the general form of the solution and the process/method to get there. Note that $\gamma$ is the Euler-Mascheroni constant.


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up vote 4 down vote accepted

Let's rewrite a little your integral :
Set $a:=2m$, $b:=\frac{2 p +1}{2}$ then

$$\int_0^\infty \frac{x^{2 m}\ln^{n}(x)}{e^{\frac{2 p +1}{2}x}}\, dx=\int_0^\infty x^a\ln^{n}(x)e^{-bx}\, dx$$ $$=\int_0^\infty \left(\frac d{da}\right)^n e^{a\ln(x)}e^{-bx}\, dx$$ $$=\left(\frac d{da}\right)^n \int_0^\infty x^a e^{-bx}\, dx$$ $$=\left(\frac d{da}\right)^n \left(b^{-a-1}\int_0^\infty t^a e^{-t}\, dt\right)$$ $$=\left(\frac d{da}\right)^n \left(b^{-a-1}\Gamma(a+1)\right)$$

A common trick to compute $\Gamma'(x)$ and derivatives is to use $\psi(x)=\log(\Gamma(x))'= \dfrac{\Gamma'(x)}{\Gamma(x)}$
(with $\psi$ the Digamma function and the derivatives the Polygamma function) so that
and so on...

Let's consider some examples of $I(n,a,b)=\left(\dfrac d{da}\right)^n \left(b^{-a-1}\Gamma(a+1)\right)$

  • $n=1$ : $$I(1,a,b)=b^{-a-1}\left[-\ln(b)+\psi(1+a)\right]\Gamma(a+1)$$ getting a generalization of your first example (case $a=2,b=\frac 32$) :
    $$I(1,2,\frac 32)=\left(\frac 32\right)^{-3}\left[-\ln\left(\frac 32\right)+\psi(3)\right]\Gamma(3)$$ with $\psi(3)=H_2-\gamma=1+\frac 12-\gamma$.

  • $n=2$ : $$I(2,a,b)=b^{-a-1}\left(\psi'(1+a) + \psi(1+a)^2 - 2\ln(b)\psi(1+a)+\ln(b)^2\right)\Gamma(a+1)$$ $\cdots$

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Did it take long to come up with this solution? Let me know how you devised it! – Pedro Tamaroff Apr 9 '12 at 23:53
@Peter: not really I must (modestly ;-) admit... When you see a $\ln(x)^n$ that seems to complicate the stuff just think that $\ln(x)$ appears with $\dfrac d{da} x^a$ ! The fun part here was that I could reuse the parameter $m$. Concerning $\Gamma$ it is of common use so that the remaining was straightforward... Cheers, – Raymond Manzoni Apr 10 '12 at 6:55

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