Natural Logarithm Integral

Consider $F(a)=\int_0^1 x^a \text{dx}=\dfrac{1}{a+1}$ differentiating with respect to a 50 times we get $F^{(50)}(a)=\int_0^1 \ln^{50}(x)x^a\text{dx}= \dfrac{50!}{(a+1)^{51}}$ put a=50 $F^{(50)}(50)=\int_0^1 \ln^{50}(x)x^{50}\text{dx}=\boxed{ \dfrac{50!}{51^{51}}}$

Differentiating under the integral sign is much slicker, but you can do this directly by integrating by parts.

Let $F(a,b) = \int_0^1 x^a (\ln x)^b \, dx.$ Then if $a,b>0,$ integrating by parts shows that $F(a,b) = -\frac{b}{a+1} F(a,b-1).$ If $b$ is a positive integer, we can apply this formula $b$ times to get $F(a,b) = (-1)^b \frac{b!}{(a+1)^b} F(a,0) = (-1)^b \frac{b!}{(a+1)^{b+1}}.$ Now plug in $a=b=50$ to get the result.

Me and a friend found a beautiful solution for this without differentiating under the integral sign.

$\int_0^1 x^{50}\ln^{50} x dx$ Substitutions: $-t=\ln x$ ; $t=-\ln x$ ; $x=e^{-t}$ ; $-dt=\frac{1}{x}dx$ ; $-e^{-t}dt=dx$ ;

Bounds: Lower: $t(0)=-\lim\limits_{x\to 0} \ln x = -(-\infty)=\infty$ ; Upper: $t(1)=\ln 1 = 0$ $\int_{\infty}^0 (e^{-t})^{50}\cdot(-t)^{50}\cdot(-e^{-t})dt = -\int_{\infty}^0 t^{50}e^{-50t}e^{-t}dt = \int_0^{\infty} t^{50}e^{-51t}dt$ Substitutions: $z=51t$ ; $\frac{1}{51}dz=dt$ ; $t=\frac{z}{51}$

Bounds: Stay the same. $\frac{1}{51}\int_0^{\infty} (\frac{z}{51})^{50}e^{-z}dz = \frac{1}{51^{51}}\int_0^{\infty} z^{50}e^{-z}dz = \frac{\Gamma(50+1)}{51^{51}} = \boxed{\frac{50!}{51^{51}}}$

Since this is an objective question, we could calculate a guess. If we use by parts then one can notice 51^51 showing up in the denominator.