Question. Solve: $$\int^1_0 \log^{2020}x dx$$
Attempt. Now I don't want any other alternative as I know there are many, I just want to know about my method: let $u=\log^{2020}x$ so that $du=2020\log^{2019}\frac{1}{x}dx$ this cancels out to a single log like the following: $$\frac{1}{2020}\int^0_{+\infty}\sqrt[2020]{u}e^{\sqrt[2020]{u}}du$$ since $\log x=\sqrt[2020]{u}$ and $x=e^{\sqrt[2020]{x}}$. Then we do a final one, $t=\sqrt[2020]{u}$ so that $dt=\frac{1}{2020}u^{-\frac{2019}{2020}}du$ hence $u=t^{2020}$ and flipping the integral to have it negative, and since the bounds don't change, we have: $$-\int^{+\infty}_0 t^{2020}e^t dt$$ and now here comes where I'm unsure, I didn't know what to do so I decided to go with the integration by parts 2020 times? As crazy as it sounds, I'd expect so from a Berkeley competition. I gave it a try:
Excluding the first term which is positive, leaving that negative out I'm pretty sure we get this?: $$-\left[t^{2020}e^t-t^{2019}e^t-t^{2018}e^t-\dots-te^t-e^t\right]^{+\infty}_0$$ and then this? $$\lim_{t\to\infty}(-t^{2020}e^t+\sum^{2019}_{k=0}e^t t^k)-1$$ (the last 1 there is because if we plug 0 we get 1 on the last term).
Again, I'm unsure whether this is even possible, but yeah, I'm stuck there and I wonder if:
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What I have done until there is correct?
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Is it possible to do it 2020 times as I did?
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Is it possible to continue from this point? Not from another point, I know other alternative methods.
Best Answer
It is useful to know Gamma Function, then realize that substituting
$$\log x=-u$$
that is
$$x=e^{-u}$$
and
$$dx=-e^{-u}du$$
your integral becomes
$$\int_0^{+\infty}u^{2020}e^{-u}du=\Gamma(2021)=2020!$$