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.
Best Answer
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 $\sin(x)$).