For what values of p does the improper integral converge?

Question

I have: $$\int^\infty_0\frac{\log(1+x)}{x^p}dx\qquad, \qquad p>0$$

I know the integral diverges if $\;0<p\le1\;$ by the comparison test, but need help with other values of $p$. Ideas? thanks!

Answer 1

We need $p>1$ to ensure integrability in a left neighbourhood of $+\infty$ and $p<2$ to ensure integrability in a right neighbourhood of the origin. If $p\in(1,2)$, the value of the integral is

$$ \int_{0}^{+\infty}\frac{\log(1+x)}{x^p}\,dx = \frac{\pi}{(p-1)\sin((p-1)\pi)} $$ by integration by parts, Euler's Beta function and the reflection formula for the $\Gamma$ function.
At $p=\frac{3}{2}$ the value of the integral is simply $2\pi$.

Answer 2

Note that we have

$$\frac{x^{1-p}}{1+x}\le \frac{\log(1+x)}{x^p}\le x^{1-p}$$

Hence, the integral $\int_0^1 \frac{\log(1+x)}{x^p}\,dx$ converges for $p<2$ and diverges elsewhere.

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In addition, we see that

$$\frac{\log(1+x)}{x^p}=x^{-p}\log(x)+x^{-p}\log\left(1+\frac1x\right)\le $$

so that

$$x^{-p}\log(x)+\frac{x^{-(p+1)}}{1+x}\le \frac{\log(1+x)}{x^p}\le x^{-p}\log(x)+x^{-(p+1)}$$

Hence, the integral $\int_1^\infty \frac{\log(1+x)}{x^p}\,dx$ converges for $1<p$ and diverges elsewhere.

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Putting it together, the integral $\int_0^\infty \frac{\log(1+x)}{x^p}\,dx$ converges for $1<p<2$ and diverges elsewhere.

Answer 3

Suppose $p>1$. We can manipulate this integral using integration by parts, with $u=\ln(1+x)$, $dv = x^{-p}$: $$\int_0^\infty \frac{\ln(1+x)}{x^p} dx = \frac{1}{1-p}\ln(1+x)x^{1-p}\bigg\vert_0^\infty - \frac{1}{1-p}\int_0^\infty \frac{1}{(1+x)x^{p-1}} dx$$ $$ = \frac{1}{p-1}\int_0^\infty \frac{1}{(1+x)x^{p-1}} dx = \frac{1}{p-1}\int_0^\infty \frac{1}{x^{p-1}+x^p} dx$$ $$\sim \frac{1}{p-1}\left(\int_0^1 \frac{1}{x^{p-1}}dx+\int_1^\infty \frac{1}{x^p} dx \right)$$ Here, $\sim$ denotes the integrals having the same behavior, as far as convergence/divergence is concerned. We can see that for $1<p<2$, both integrals converge, but for $p\geq 2$, the first integral fails to converge. You can get the answer more rigorously using the usual arguments.

Answer 4

Near $0^+$, the integrand is equivalent to $$\frac {1}{x^{p-1}} $$ it converges if $p-1 <1$.

Near $+\infty $, if we put $x+1=t $, the new integrand is equivalent to

$$\frac {ln (t)}{t^p} $$ it is Bertrand type which converges if $p>1$

finally, the integral converges $\iff 1 <p <2$.