# For what real values of a does the integral $\int\limits_0^1(-\ln x)^a\,dx$ converge?

So far I've substituted $$t = -\ln x,\quad x = e^{-t} \, dt = -\frac 1 x\,dx$$ Now I need to deal with the integral of $$\int_0^\infty t^ae^{-t} \, dt$$ which I've split into parts: $$\int_0^1 t^ae^{-t} \, dt+ \int_1^\infty t^ae^{-t} \, dt$$

It's pretty easy to see that the integral from $0$ to $1$ converges with $a>-1$. For the second part, the exponent beats out the polynomial as $t$ tends to infinity, so it looks like the integral converges for all $a$, but how do I formally justify this?

Thanks for your time.

## 1 Answer

Define $$f(a) = \int_{0}^{\infty}t^a e^{-t}\,\mathrm{d}t$$ We easily see $f(1) = 1$. Now, using integration by parts, we find $$f(a) = \int_{0}^{\infty}t^a e^{-t}\,\mathrm{d}t = 0 + a\int_{0}^{\infty}t^{a-1}e^{-t}\,\mathrm{d}t = af(a - 1)$$

Thus we have $f(a + 1) = af(a)$. By induction, $f(a) = a!$, and this is in fact the definition of the well known Euler gamma function.

To formally prove convergence, use the limit comparison test with (for example) $f(t) = t^2$. Thus we have $$\lim_{t \to \infty}t^{a + 2}e^{-t} = 0$$ and thus the integral converges absolutely for $a > 0$.