# gamma and beta function....

Show that $$\int_0^1 x^m (\ln \frac{1}{x})^n\mathrm dx=\frac{\Gamma(n+1)}{(m+1)^{n+1}}$$

Let, $$\ln\frac{1}{x}=z$$ . So, $$x=e^{-z}$$ and, $$\mathrm dx =-e^{-z}\mathrm dz$$ \begin{align} \int_0^1 x^m (\ln \frac{1}{x})^n\mathrm dx&= \int^\infty_0 (e^{-z})^mz^n(-e^{-z}\mathrm dz)\\&=\int^\infty_0 e^{-(m+1)z}z^{(n+1)-1} \mathrm dz\end{align}

In next line, they wrote that

$$\frac{\Gamma(n+1)}{(m+1)^{n+1}}$$

How did they found $$(m+1)^{n+1}$$? I know that

$$\Gamma(n+1)=\int_0^\infty e^{-z}z^{n+1 -1}\mathrm dz$$

How did they convert $$e^{-(m+1)z}$$?

$$=\int_0^\infty e^t (\frac{t}{m+1})^{(n+1)-1}\mathrm dt$$ $$=\frac{1}{(m+1)^{(n+1)-1} \int_0^\infty e^{-t}t^{n+1-1}\mathrm dt}$$ $$=\frac{\Gamma(n+1)}{(m+1)^{n}}$$

But, I got $$n$$ in denominator's power. They wrote $$n+1$$

• its not \infinity it is \infty Jul 31 at 6:14
• Use substitution $(m+1)z=t$ Jul 31 at 6:17
• @PNDas Could pls check my work?>
– user954149
Jul 31 at 6:28

When you substitute $$(m+1)z = t$$, you get
$$e^{-(m+1)z} = e^{-t}, \ z = \cfrac{t}{m+1}, \ dz = \cfrac{1}{m+1} dt$$
So, $$\displaystyle \int^\infty_0 e^{-(m+1)z}z^{(n+1)-1} \ dz =$$
$$\displaystyle \int_0^{\infty} e^{-t} \left(\cfrac{t}{m+1}\right)^{(n+1 - 1)} \cfrac{1}{m+1} \ dt$$
$$\displaystyle = \int_0^{\infty} e^{-t} t^{(n+1 - 1)} \cfrac{1}{(m+1)^{n+1}} \ dt$$
$$\displaystyle = \frac{\Gamma(n+1)}{(m+1)^{n+1}}$$