# Gamma fuction and integral

• Dec 10th 2009, 07:50 PM
osodud
Gamma fuction and integral
Hello.

I just cant solve this problem. Who can do it?

Using gamma fuction calculate the following integral:
$\displaystyle \int_o^1 x^3 [ln(\frac{1}{x})]^3 dx$

Thanks a lotĦĦ
• Dec 10th 2009, 08:28 PM
Chris L T521
Quote:

Originally Posted by osodud
Hello.

I just cant solve this problem. Who can do it?

Using gamma fuction calculate the following integral:
$\displaystyle \int_o^1 x^3 [ln(\frac{1}{x})]^3 dx$

Thanks a lotĦĦ

Let $\displaystyle u=\ln\!\left(\tfrac{1}{x}\right)\implies \,du=-\tfrac{1}{x}\,dx$. Note that $\displaystyle u=\ln\!\left(\tfrac{1}{x}\right)\implies e^{-u}=x$

Also note that $\displaystyle u(0)\rightarrow\infty$ and $\displaystyle u(1)=0$.

Therefore, $\displaystyle \int_0^1x^3\left[\ln\!\left(\tfrac{1}{x}\right)\right]^3\,dx\xrightarrow{u=\ln\!\left(\tfrac{1}{x}\right )}{} \int_{\infty}^{0}-x^4 u^3\,du=\int_0^{\infty}e^{-4u}u^3\,du$

Now make the substitution $\displaystyle t=4u\implies \,dt=4\,du$. Also note that $\displaystyle t(0)=0$ and $\displaystyle t(\infty)=\infty$

Therefore $\displaystyle \int_0^{\infty}e^{-4u}u^3\,du\xrightarrow{t=4u}{}\tfrac{1}{4}\int_0^{ \infty}e^{-t}\left(\tfrac{t}{4}\right)^{4}\,dt=\tfrac{1}{1024 }\int_0^{\infty}e^{-t}t^{4}\,dt$

Now apply the definition of the gamma function to get $\displaystyle \tfrac{1}{1024}\int_0^{\infty}e^{-t}t^{4}\,dt=\frac{\Gamma\!\left(5\right)}{1024}=\f rac{24}{1024}=\frac{3}{128}$

Does this make sense?