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How do I determine whether the integral $$\int^{\infty}_1\frac{\sin x}{(\ln(x+1)-\ln x)^a} \mathrm{dx}$$ converges absolutely or not and for which $a$?

I actually to be honest have no idea how to approach this problem and could use some hints on how to solve this. So I took a look at this question, and I think I can use the comparison test knowing that the integrand $$\int^{\infty}_1\biggl|\frac{\sin x}{x}\biggr|\mathrm{dx}$$ diverges.

My attempt:

Proving that $$\int^{\infty}_1\biggl|\frac{\sin x}{(\ln(x+1)-\ln x)^a} \biggr|\mathrm{dx}$$ diverges:

Knowing that $$\frac{\sin(x)}{(\ln(x+1)-\ln x)^a}>\frac{\sin(x)}{x}$$ for $x\in [1,\infty)$ and $\forall a>0$, we can deduce that since $$\int^{\infty}_1\biggl|\frac{\sin x}{x}\biggr|\mathrm{dx}$$ diverges it follows that $\displaystyle\int^{\infty}_1\biggl|\frac{\sin x}{(\ln(x+1)-\ln x)^a} \biggr|\mathrm{dx}$ diverges

As per Toby Mak's comments, I will be analyzing the integrand $\int^{\infty}_1\sin(x)x^a\mathrm{dx}$. This integral, for all $a\geq 0$ diverges. Hence our integrand also diverges.

  • $\int_1^{\infty} \frac{\sin x}{x} \ dx$ (without the absolute value sign) converges, as the post you linked says. The absolute value signs are another case. – Toby Mak Jul 12 '21 at 08:15
  • @TobyMak I proved that it doesn't converge for the absolute value. How do I prove it converges or diverges without the absolute sign? –  Jul 12 '21 at 08:19
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    When $x$ is large, $\ln(x+1) - \ln x = \ln \left( \frac{x+1}{x} \right) = \ln (1 + 1/x) \approx 1/x$ from the Taylor series. So compare the integrand to $\frac{\sin x}{(1/x)^a}$: since their ratio as $x$ tends to infinity is $1$, by the limit comparison test either both converge or both diverge. – Toby Mak Jul 12 '21 at 08:24
  • And see this post for the convergence of $\int \frac{\sin x}{x^a} \ dx$. – Toby Mak Jul 12 '21 at 08:29
  • @TobyMak : I think your argument about the limit comparison test works for absolute convergence, but not for convergence. – FiMePr Jul 12 '21 at 10:09
  • @FiMePr True, conditional convergence is a different story. – Toby Mak Jul 12 '21 at 10:53
  • @TobyMak but what you said in your first comment is not true I think. The question's answer which I cited proves that the integrand diverges, hence my written answer is correct? –  Jul 12 '21 at 13:20
  • @TobyMak also, I did what you said in my edit –  Jul 12 '21 at 13:26

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It is not a bad idea to use substitution $u=1+\frac1{x}$ then it is all a little bit more obvious

$$\int_{1}^{\infty} \frac{\sin(x)}{\ln(1+\frac{1}{x})^a} \, dx = \int_{2}^{1} \frac{\sin(\frac1{u-1})}{2\ln(u)^a(u-1)^2} \, du $$

is not convergent.

From $1$ to $2$, it is: $0 \leq \ln(u) \leq u-1$ and $\frac1{(u-1)^a} > 1$ as well so

$$\left | \int_{2}^{1} \frac{\sin(\frac1{u-1})}{2\ln(u)^a(u-1)^2} \, du \right | \geq \left | \int_{2}^{1} \frac{\sin(\frac1{u-1})}{2(u-1)^a(u-1)^2} \, du \right | \geq \left | \int_{2}^{1} \frac{\sin(\frac1{u-1})}{2(u-1)^2} \, du \right |$$

This is not convergent because this is not convergent:

$$\int_{2}^{1} \frac{\sin(\frac1{u-1})}{(u-1)^2} \, du = -\cos(\frac{1}{u-1}) \Big |_{2}^{1}=-\cos(x) \Big |_{1}^{\infty}$$

So the initial integral is not convergent for any $a \geq 0$. (For $a=0$ the divergence is coming from the divergence of infinite integral of $\sin(x)$).