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I have two integrals to decide and prove if they converge or diverge, I solved them (or tried), but I'm unsure of my solution and got some questions:

$1)$ $$\int_1^3\frac {\sin(1-x)}{(x-1)^2}\,dx $$ What I tried to do: I said that this integral is less than $$\int_1^3\frac{1}{(x-1)^2}\,dx,$$ and if I substitute $t= x-1$, I will get $$\int_0^2\frac {1}{t^2}\,dt,$$ which diverges. Now I know that I can't say that my integral diverges using this fact, sadly I couldn't finish it.

$2)$ $$\int_1^\infty\frac{\ln(1+e^x)-x}{x^2}\,dx$$

What I did: I found the limit $$\lim_{x\to\infty}\frac {\ln(1+e^x)}{x} = 1,$$ so I decided to take the function $\frac {1}{x}$ (I checked the limit with my integrand and got $1$), and $\frac {1}{x}$ diverges so my integral has to diverge.

I feel like I went a little far, I tried a couple functions, and most of them failed or weren't good for comparison test, I would really appreciate any explanation and if someone can approve my answers or point out my mistakes.
Thanks in advance.

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  • $\begingroup$ I think that the first one diverges for a different reason, namely that $\sin x\to x$ when $x\to 0$. I think the second one should converge, you probably won't get $\frac 1x$ if you consider the whole numerator... $\endgroup$
    – abiessu
    Jan 20 at 21:49
  • $\begingroup$ More specific for the second one: note that $$\ln(1+e^x)-x=\ln\left(\frac{1+e^x}{e^x}\right)$$ $\endgroup$
    – abiessu
    Jan 20 at 21:58
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    $\begingroup$ $\int_1^3\frac{1}{(x-1)^2}\,dx$ diverges and $\frac{\sin (1-x)}{(x-1)^2}\le \frac{1}{(x-1)^2}$ means that $\frac{\sin (1-x)}{(x-1)^2}$ could converge. Your explanation of the first integral divergence is not valid. $\endgroup$
    – Raffaele
    Jan 20 at 21:58
  • $\begingroup$ Note that $ \sin x \geqslant \frac{2x}{\pi}$ and, thus, $ \frac{\sin x}{x^2} \geqslant \frac{2}{\pi x}$ for $0 \leqslant x \leqslant \frac{\pi}{2}$ . $\endgroup$
    – RRL
    Jan 20 at 22:24
  • $\begingroup$ Also $ \int_1^3 \frac{\sin(1-x)}{(x-1)^2} \,dx = -\int_0^{\pi/2}\frac{\sin x}{x^2}\, dx - \int_{\pi/2}^{2}\frac{\sin x}{x^2}\, dx$. $\endgroup$
    – RRL
    Jan 20 at 22:24
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For the first integral notice that

$$\frac{\sin(1-x)}{(x-1)^2} = \frac{\sin(-(x-1))}{(x-1)^2} = -\frac{\sin(x-1)}{(x-1)^2} \sim -\frac{1}{x-1} ~~~~~ \text{as} ~~~~~ x\to 1$$

Hence the first integral does diverge.

About the second one we have

$$\frac{\ln(1 + e^x) - x}{x^2} \to 0$$

as $x\to +\infty$ hence the second integral does converge.

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