Define $\alpha$, such that the following integrals converges: $$I_1 = \int_{0}^{\infty} (x + \frac{1}{x})^{\alpha} \ln (1+x^{- 3 \alpha}) d x, \\ I_2 = \int_0^1 x^{\alpha}(1-x)^{\beta} \ln x d x , \\ I_3 = \int_0^{\infty} \frac{|\sin x|}{e^{x^2 \sin^2 (x)}} d x, \\ I_4 = \int_0^{\infty}\frac{d x}{1 + x^{\alpha} \sin x^2}. $$ May you suggest some ideas, please? Can't solve them for about a week...

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    $\begingroup$ One integral at a time please. $\endgroup$
    – zhw.
    Mar 9, 2017 at 23:38
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    $\begingroup$ Your 3rd integral does not contain any $\alpha$ , although its convergence as written is an interesting problem. Your second integral contains both $\alpha$ and $\beta$, but it is closely related to a Beta function integral so the question of convergence will be "easy" (LOL). $\endgroup$ Mar 9, 2017 at 23:44
  • $\begingroup$ I'm voting to close this question as off-topic because this question is a collection of more than one "how to compute this integral" and would be better off asked as separate questions with work shown on where you are stuck. $\endgroup$
    – R_D
    Mar 11, 2017 at 5:26

1 Answer 1


The first two:

$I_1$: from $\ln(1+x)\leq x$, $$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\leq \left(x+\frac1x\right)^\alpha \,\frac1{x^{3\alpha}}=\left(1+\frac1{x^2}\right)^\alpha\,\frac{1}{x^{2\alpha}}, $$ so convergence of the integral at $\infty$ is guaranteed if $2\alpha>1$, i.e. $\alpha>1/2$. For $\alpha\leq1/2$, since we only care about big values of $x$ we have $$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\geq x^\alpha\,\frac1{2x^{3\alpha}}=\frac1{2x^{2\alpha}} $$ and the integral will diverge.

At zero, $$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\geq -3\alpha\,\frac {\ln x}{x^\alpha} $$ and the integral will diverge if $\alpha\geq1$. For $\alpha <1$, $$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\leq 2^\alpha\,\frac {\ln2-3\alpha\ln x}{x^\alpha}$$

$I_2$: we have, for any $\alpha\geq0$, $$ \left|\int_0^1 x^\alpha(1-x)^{\beta}\,\ln x\,dx\right|\leq\int_0^1|\ln x|\,dx=1, $$ so the integral converges. For $-1<\alpha<0$, we have $$ x^{\alpha}\ln x=x^{\alpha-(\alpha+1)/2}(x^{(\alpha+1)/2}\ln x), $$ the expression in brackets will be bounded, and the integral converges because $-1<(\alpha-1)/2<0$. For $\alpha\leq-1$ the integral will diverge. The constant $\beta$ deals with the convergence at $1$. When $\beta\geq0$, the integral converges as given by $\alpha$ above. For $\beta<0$, we can see the behaviour by changing variables $x\longmapsto 1-x$, so $$ I_2=\int_0^1 (1-x)^\alpha x^\beta\,\ln(1-x)\,dx. $$ Close to $0$, $$(1-x)^\alpha x^\beta\,\ln(1-x)\simeq x^\beta(-x)=-x^{\beta+1}.$$ So the integral will converge when $\beta+1>-1$, that is $\beta>-2$. It will diverge for $\beta\leq-2$.

  • $\begingroup$ Your answer to the second integral assumes that $\beta \geq 0$. It is unclear to me whether $\beta > -1$ is the condition on $\beta$, irrespective of the value of $\alpha$. (It would be clear if $\ln x$ behaved similarly at $x=1$ to $x=0$ but it does not.) Still, I am pretty sure the integral converges if and only if $\alpha$ and $\beta$ are both greater than $-1$. $\endgroup$ Mar 10, 2017 at 0:00
  • $\begingroup$ You are totally right. $\endgroup$ Mar 10, 2017 at 1:43
  • $\begingroup$ @Mark: I have included the case for $\beta$. $\endgroup$ Mar 10, 2017 at 4:20
  • $\begingroup$ There is a mistake in the analysis of $I_1$. You can wolfram it and find out that it does not converge at $\alpha > 1$. $\endgroup$
    – Kamil
    Mar 10, 2017 at 18:50
  • $\begingroup$ Good point, thanks. I completely missed the analysis at zero. The integral ideed diverges for $\alpha\geq1$, I have included that. $\endgroup$ Mar 10, 2017 at 20:51

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