Gamma fuction and integral

Quote:

Originally Posted by osodud

Hello.

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

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

Thanks a lot¡¡

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

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

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

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

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

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

Does this make sense?