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


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 $n+1$

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

$\displaystyle = \int_0^{\infty} e^{-t} t^{(n+1 - 1)} \cfrac{1}{(m+1)^{n+1}} \ dt$

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


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