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 Post subject: Advanced Math Integral
PostPosted: Tue, 14 Oct 2008 00:38:54 UTC 
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Joined: Thu, 11 Sep 2008 01:59:14 UTC
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[Integral from 0 to infinity] of [ln(x) * e^(-3x)]

Have fun!

Hint: Exponential integral Ei(x)


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PostPosted: Sun, 29 Mar 2009 01:31:20 UTC 
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Joined: Sat, 21 Feb 2009 21:40:14 UTC
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Location: ncsu soph
(I adopt the convention \log z = \ln z).

No need to use Ei. Just use the first Euler-Mascheroni integral -\gamma=\int_0^{\infty}e^{-t}\log t\;\mathrm{d}t, which is proven as follows:

\partial_{z}\Gamma(z)=\displaystyle\int_0^{\infty}\partial_{z}e^{(z-1)\log t}e^{-t}\mathrm{d}t=\int_0^{\infty}t^{z-1}e^{-t}\log t\;\mathrm{d}t from the integral definition of the gamma function, so as z goes to 1

\Gamma^{\prime}(1)=\displaystyle\int_0^{\infty}e^{-t}\log t\;\mathrm{d}t. Using the Weierstrauss product \Gamma(z+1)=\displaystyle e^{-\gamma z}\prod_{n=1}^{\infty}\frac{e^{z/n}}{1+\frac{z}{n}} and logarithmically differentiating

\frac{\Gamma^{\prime}(z+1)}{\Gamma(z+1)}=-\gamma+\displaystyle\sum_{n=1}^{\infty}\frac{z}{n(n+z)}. Simply allowing z to go to 0 gives \Gamma^{\prime}(1)=-\gamma, \gamma=\displaystyle\lim_{\zeta\to\infty}\log\cfrac{1}{\zeta}+\sum_{n=1}^{\zeta}\frac{1}{n} being the Euler-Mascheroni constant. So, -\gamma=\displaystyle\int_0^{\infty}e^{-t}\log t\;\mathrm{d}t.

Now consider integrals of the form \displaystyle I(n)=\int_0^{\infty}e^{-nz}\log z\;\mathrm{d}z; substitute \mathrm{d}z=\cfrac{\mathrm{d}x}{n}, giving \displaystyle I(n)=\frac{1}{n}\int_0^{\infty}e^{-x}\log\cfrac{x}{n}\;\mathrm{d}x=\frac{1}{n}\left(\int_0^{\infty}e^{-x}\log x\;\mathrm{d}x-\log n\cdot\int_0^{\infty}e^{-x}\mathrm{d}x\right) which after evaluating and simplifying is

\displaystyle I(n)=-\frac{1}{n}\left(\gamma+\log n) which is the solution for all n>0. To answer your question specifically, \displaystyle I(3)=\int_0^{\infty}e^{-3x}\log x\;\mathrm{d}x=-\frac{1}{3}\left(\gamma+\log 3).


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