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

Question

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.

Answer 1

Hint. Following your approach by differentiation under the integral sign, note that $$\int_{0}^{1} x^t (\ln(x))^n \, dx=\frac{d^n}{dt^n}\left( \int_0^1 x^t dx\right)=\frac{d^n}{dt^n} \left((t+1)^{-1}\right).$$

Answer 2

$$ \int_{0}^{1}\! \left( x\ln \left( x \right) \right) ^{n}\,{\rm d}x={ \frac { \left( -1 \right) ^{n}n!}{ \left( n+1 \right) ^{n+1}}} $$

Answer 3

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.

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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 $n=50$.

Answer 4

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 $$\displaystyle I=+\frac{a!}{(a-a)!}u^{a-a}\cdot\left(\frac1{a+1}\right)^{a+1}e^{(a+1)u}=\frac{a!}{(a+1)^{a+1}}=\frac{50!}{51^{51}}.$$

Answer 5

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,

$$\frac{50!}{51^{50}}\int_0^1 x^{50}dx.$$

Answer 6

$$ \int_{0}^{1}\left(x\ln(x)\right)^{50} dx $$

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: $\Gamma\left(n+1\right) = n!$

$$ =\frac{50!}{51^{51}} $$