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find values of n for which the integral converges

$ \int_0^1 (\ln\left(\frac{1}{x}\right))^n $ dx
i am able to understand that Integral needs to be broken in two parts so that convergence can be checked at x=1 and x=0 separately.

for the first part if n<0 then its proper integral similarly for the second half n>0 then its proper integral

i need to check for the other remaining cases but unable to get how?

Svidi Runs
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  • You can recall the integral is $n!$ – Тyma Gaidash Feb 23 '24 at 23:24
  • can you link me to any hint – Svidi Runs Feb 23 '24 at 23:28
  • What are you trying to do? Evaluate it, or prove it converges? This thing is continuous at $1$ so the only possible issue could be near $x=0$, but you can use the fact that $h^s\ln(1/h)^n\to0$ as $h\to0$ for arbitrary powers $s>0$. In particular, think about a choice $0<s<1$. Alternatively make variable substitution $x=e^{-t}$ which (i) leads to an exact evaluation and (ii) more clearly shows convergence – FShrike Feb 23 '24 at 23:47
  • What are $\ln(a^n)$ and $\ln(1/a)$? – Ted Shifrin Feb 23 '24 at 23:47
  • @FShrike for negative n , x=1 is also point of infinite discontinuity, I'm trying to prove convergence. – Svidi Runs Feb 24 '24 at 00:55
  • Ok, for negative $n$ this thing does not converge. I was thinking about $n\ge0$. You've had some comments already, what have you tried? – FShrike Feb 24 '24 at 01:01
  • Use properties of the logarithm to simplify the expression. – CyclotomicField Feb 24 '24 at 01:23

1 Answers1

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$\int _0^1(\ln\frac{1}{x})^n\mathrm{d}x\overset{t=\frac{1}{x}}{=}\int_1^{+\infty}(\ln t)^nt^{-2}\mathrm{d}t\overset{z=\ln t}{=}\int_0^{+\infty}z^n\mathrm{e}^{-z}\mathrm{d}z=\Gamma(n+1)=n!$

RainField
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