Discuss the convergence of $\int_0^1x^n \left[\log({1\over x})\right]^m \, dx$

Question

Discuss the convergence of

$$\int_0^1x^n\left[\log\left({1\over x}\right)\right]^m \, dx$$

Need some clues. I know that both $0$ and $1$ are points of discontinuities.

Answer 1

Hint: It is easier to look at

$$I = \int_{1}^{\infty} \frac{(\ln(y))^m}{y^{n+2}}, $$

by making the change of variables $y=1/x$.

Answer 2

The integral is the same as $$ (-1)^m \int_0^1 x^n (\log x)^m \, dx. $$

If $u=\log x$ then $du = dx/x$ and $x = e^u$. Then as $x\downarrow0$ we have $u\to-\infty$ and when $x=1$ then $u=0$. And $x^n\,dx= x^{n+1}\,(dx/x)$.

So the integral becomes $$ (-1)^m \int_{-\infty}^0 e^{(n+1)u} u^m \, du. $$

It may be convenient to write $w = (n+1)u$ so we get $$ (-1)^m \int_{-\infty}^0 e^w \left(\frac{w}{n+1}\right)^m \, \frac{dw}{n+1} = \frac{(-1)^m}{(n+1)^{m+1}} \int_{-\infty}^0 e^w w^m \, dw. \tag 1 $$

Now a big hint: $$ \int w^m\Big(e^w\,dw\Big) = \int s\,dt. $$ Integration by parts only reduces the power of $w$ from $m$ to $m-1$; doing it again reduces it to $m-2$, and so on, so you'll get $m(m-1)(m-2)\cdots$, in other words, a factorial. In the last step, you'll have $$ \Big[\cdots\cdots\cdots\Big]_{-\infty}^0 - (\cdots\cdots)\int\cdots\cdots\cdots. $$ The part in square brackets can be handled by L'Hopital's rule.

If you only want to know about convergence rather than about the value of the integral, think about the very last expression in $(1)$.