Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks.

share|improve this question
add comment

1 Answer

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
$\Gamma'(x)=\psi(x)\Gamma(x)$,
$\Gamma''(x)=\left(\psi'(x)+\psi(x)^2\right)\Gamma(x)$
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 $\psi(3)=H_2-\gamma=1+\frac 12-\gamma$.

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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