gamma functionimproper-integralsintegration
I have to evaluate $\int_0^1 x^2 \ln(\frac1x)^3 $.I used gamma function and used substitution $t=\ln (\frac {1}{x})^3$.
In this I get to integrate from $1$ to $-\infty$ with a minus sign outside.Because of this minus sign by interchanging upper and lower limit
I get to integrate from $+\infty$ to $1$.
Since $x>0$ I took this integration from 0 to 1.
Can I make that change of $+\infty$ => 0.
The answer I got is $\Gamma(2)=1$.Is this correct.Any help is appreciated
Summarising the comments, after substitution of $u=ax^2$ and $du=2axdx$, you'll get $$ \int_0^\infty x^2 e^{-ax^2}dx=\frac{1}{2a^{3/2}}\int_0^\infty u^{3/2-1}e^{-u} du=\frac{\Gamma(3/2)}{2a^{3/2}}=\frac{\sqrt{\pi}}{4a^{3/2}}. $$
The result (below) as expected includes $ \frac{\Gamma(1/6)\Gamma(1/2)}{\Gamma(2/3)} = \sqrt{\pi}\frac{\Gamma(1/6)}{\Gamma(2/3)}$ and an hypergeometric function.
See Eq.(9) in : http://mathworld.wolfram.com/AppellHypergeometricFunction.html
For some particular values of $b$, the hypergeometric function reduces to functions of lower level.
Best Answer
Let me solve it from the first point. If I got correctly you are working on $$-\int_0^1x^2\ln^3(x)dx$$ Letting $x=\text{e}^{-y}$, the integral becomes $$(-1)^{4}\int_0^{\infty}y^3\text{e}^{-3y}dy$$ Now if we set $3y=u$, the latter integral becomes $$\int_0^{\infty}\frac{u^3}{3^3}\text{e}^{-u}\frac{du}{3}=\frac{1}{3^4}\Gamma(4)$$