Evaluate $\int_0^1(x\ln(x))^{50}dx$ 
Evaluate $$\int_0^1(x\ln(x))^{50} dx.$$

Here are my steps so far using differentiation under the integral sign:
$$I(t) = \int_0^1(x\ln(x))^t dx$$
$$I'(t) = \frac d{dt}\int_0^1(x\ln(x))^t dx = \int_0^1\frac \partial{\partial t}(x\ln(x))^t dx = \int_0^1(x\ln(x))^t\ln(x\ln(x)) dx$$
I can't find a way to continue so hints are appreciated.
 A: Hint:
By parts, 
$$\int_0^1x^{50}\log^{50}xdx=\left.\frac{x^{51}}{51}\log^{50}x\right|_0^1-\frac{50}{51}\int_0^1x^{50}\log x^{49}dx.$$
The first term evaluates to $0$, and the second shows an easy recurrence.
After $50$ iterations,
$$\frac{50!}{51^{50}}\int_0^1 x^{50}dx.$$
A: Hint. Following your approach by differentiation under the integral sign, note that
$$\int_{0}^{1} x^t (\ln(x))^n \, dx=\frac{d^n}{dt^n}\left( \int_0^1 x^t dx\right)=\frac{d^n}{dt^n} \left((t+1)^{-1}\right).$$
A: 
I thought it might be instructive to present an approach that is efficient, elementary, and circumvents differentiation under the integral.  To that end, we proceed.


The integral can be evaluated straightforwardly by enforcing the substitution $x\mapsto e^{-x/(n+1)}$.  Using this substitution, we have for any integer $n\ge 0$
$$\begin{align}
\int_0^1 x^n\log^n(x)\,dx&=\frac{(-1)^n}{(n+1)^{n+1}}\underbrace{\int_0^\infty x^ne^{-x}\,dx}_{n!}\\\\
&=\frac{(-1)^nn!}{(n+1)^{n+1}}
\end{align}$$
Then, let $n=50$.
A: I am going to use tabular integration for a different approach to this problem. It is perhaps not quite as elegant as the Leibniz integral rule, but I think it is still quite interesting.
We have $I=\displaystyle\int_0^1 \left(x\ln x\right)^{50}\,\mathrm dx$.
Letting $u=\ln x$, we have that $\mathrm dx=e^u\,\mathrm du$ and $I=\displaystyle\int_{-\infty}^0u^{50}e^{51u}\,\mathrm du$.
For the sake of transparency about what is happening, let $a=50$, such that
$I=\displaystyle\int_{-\infty}^0u^ae^{(a+1)u}\,\mathrm du$.
Using tabular integration:
$$\require{cancel}\begin{array}{c|c|c}
+&u^a&\cancel{e^{(a+1)u}} \\
-&au^{a-1}&\frac1{a+1}e^{(a+1)u} \\
+&\frac{a!}{(a-2)!}u^{a-2}&\left(\frac1{a+1}\right)^2e^{(a+1)u} \\
-&\frac{a!}{(a-3)!}u^{a-3}&\left(\frac1{a+1}\right)^3e^{(a+1)u} \\
\vdots&\vdots&\vdots\\
+&\frac{a!}{(a-a)!}u^{a-a}&\left(\frac1{a+1}\right)^ae^{(a+1)u} \\
&&\left(\frac1{a+1}\right)^{a+1}e^{(a+1)u}
\end{array}
$$
In this process, the desired terms are found by going across from the first column to the second, then diagonally down to the third column. For instance, the first term is $+u^a\cdot\frac1{a+1}e^{(a+1)u}$. Note that the sign on the left alternates, and it will be positive for the last term because $a$ is even.
Normally, we would simply multiply together all of the terms and then evaluate, but this enormous integral would quickly become unwieldy. Hence, we must take a more clever approach.
Note that, with the $u$-substitution, the new bounds of integration are $-\infty$ and $0$.
When $u$ approaches $-\infty$, all of the terms on the right side approach $0$ quicker than the left side approaches $-\infty$, so each of those terms vanishes. (The proof is left as an exercise to the reader).
At $u=0$, all terms on the left except for the very last one will vanish. This last term, the only one that remains, will be $$\displaystyle I=+\frac{a!}{(a-a)!}u^{a-a}\cdot\left(\frac1{a+1}\right)^{a+1}e^{(a+1)u}=\frac{a!}{(a+1)^{a+1}}=\frac{50!}{51^{51}}.$$
A: $$
\int_{0}^{1}\left(x\ln(x)\right)^{50} dx
$$
Make substitutions:
$T=-\ln(x)$, $-T=\ln(x)$
$x=e^{-T}$
$-e^{-T} dT = dx$
$-dT=\frac{1}{x}dx$
Upper limit: $T(1) = \ln(1) = 0$
Lower limit: $T(0)=-\lim_{x \to 0} \ln(x) = \infty$
$$
=\int _{\infty \:}^0\left(e^{-T}\right)^{50}\left(-T\right)^{50}e^{-T}dT
$$
$$
=-\int _0^{\infty \:}\left(e^{-T}\right)^{50}\left(-T\right)^{50}e^{-T}dT
$$
$$
=\int _0^{\infty \:\:}T^{50}e^{-51}dT
$$
More substitutions:
$n = 51T$
$\frac{1}{51}dn=dT$
$T=\frac{1}{51}n$
$$
\frac{1}{51}\int _0^{\infty }\left(\frac{n}{51}\right)^{50}e^{-n}dn\:
$$
$$
\frac{1}{51^{51}}\int _0^{\infty }n^{50}e^{-n}dn
$$
$$
=\frac{\Gamma \left(50+1\right)}{51^{51}}
$$
Knowing that: $\Gamma\left(n+1\right) = n!$
$$
=\frac{50!}{51^{51}}
$$
A: $$
\int_{0}^{1}\! \left( x\ln  \left( x \right)  \right) ^{n}\,{\rm d}x={
\frac { \left( -1 \right) ^{n}n!}{ \left( n+1 \right) ^{n+1}}}
$$
