Evaluating $\int _0 ^1 x^k(\ln x)^m dx$ for integer $k$ and $m$

Question

Find a formula for $\displaystyle \int _0 ^1 x^k(\ln x)^m dx$ that works for all positive integers $k$ and $m$. Use integration by parts $m$ times with $k$ fixed.

Answer 1

Here is a start, use the change of variables $\ln(x)=-t$

$$ \displaystyle \int _0 ^1 x^k(\ln x)^m dx =\left( -1 \right) ^{m}\int _{0}^{\infty }\! {t}^{m}{{\rm e}^{-t \left( k+1\right) }}{dt}.$$

Now, you need another change of variables to relate the last integral to gamma function. I let you do it. Here is the final answer

$$ \left( -1 \right)^{m}\frac{ \Gamma(m+1) }{ (k+1)^{m+1} }=\left( -1 \right)^{m}\frac{ m! }{ (k+1)^{m+1} } . $$

Added: The other change of variables you need is $ (k+1)t=u $. Use it and compare with the gamma function.

Answer 2

Integrating by parts works out nicely:

Denote $$\varphi(k,m)=\int_0^1x^m(-\log x)^kdx$$

Note we need the minus on the $\log$, since it is negative over $[0,1]$!

Now take $$u=(-\log x)^k$$ $$dv=x^m$$ We get $$u'=-k(-\log x)^{k-1}\frac{1}{x}$$ $$v=\frac{x^{m+1}}{m+1}$$

Thus IVP says

$$\varphi(k,m)=\left.(-\log x)^k\frac{x^{m+1}}{m+1}\right|_0^1+\frac{k}{m+1}\int_0^1(-\log x)^{k-1}\frac{x^{m+1}}{x}dx$$

This gives a nice recurrence, after noting both limits on the first term on the right hand side vanish:

$$\varphi(k,m)= \frac{k}{m+1}\int_0^1(-\log x)^{k-1}{x^{m}}dx$$

that is $$\varphi(k,m)=\frac{k}{m+1}\varphi(k-1,m)$$

But then $$\varphi(k,m)=\frac{k!}{(m+1)^{k}}\varphi(0,m)$$

Thus, all we need to is to know what $$\varphi(0,m)=\int_0^1x^m dx$$ is. But you can do that, yes?

Answer 3

Here is a different approach using exponential generating series. Notice that $$\sum_{m=0}^{\infty}\frac{z^{m}}{m!}\left(\int_{0}^{1}x^{k}\left(\log x\right)^{m}dx\right)=\int_{0}^{1}x^{k}\left(\sum_{m=0}^{\infty}\frac{z^{m}}{m!}\left(\log x\right)^{m}\right)dx$$ $$=\int_{0}^{1}x^{k+z}dx=\frac{1}{z+k+1},$$ and the right hand side has the expansion

$$\frac{1}{k+1}\cdot\frac{1}{1+\frac{z}{k+1}}=\frac{1}{k+1}\sum_{j=0}^{\infty}\frac{(-1)^{j}}{\left(k+1\right)^{j}}z^{j}.$$ Comparing coefficients, we find that $$\int_{0}^{1}x^{k}\left(\log x\right)^{m}dx=\frac{m!(-1)^{m}}{(k+1)^{m+1}}.$$

Answer 4

Considering the function $f(a)=\int_0^1x^adx=\frac{1}{a+1}$ now doing differentiation under integral sign of $f(a)$ wrt $a$, $m$ times we have $f^{m}(a)=\int_0^1x^a(lnx)^mdx=\frac{(-1)^mm!}{(a+1)^{m+1}}$ now put $a=k$ we get $f^{m}(k)=\int_0^1x^k(lnx)^mdx=\frac{(-1)^mm!}{(k+1)^{m+1}}$