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$$\int_e^\infty \frac{\ln x\cdot \cos(x^2)}{(x+1)^{\frac 3 2}} \, dx$$

After an hour of trying to compare this expression to something, I've put it in WolframAlpha which calculated it precisely so basically that means this integral converges(I think?), but I've failed to prove it. I've tried limiting $cos({x^2})$ with $-1$ and $1$, tried looking at the absolute convergence.

Everything that I found bigger than this always seems to diverge, and anything smaller seems to converge or go to $-\infty$ as $x$ goes to $+\infty$.

Any hints would be appreciated, thank you in advance!

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$$\int_e^\infty \frac{\ln(x)\cos(x^2)}{(x+1)^{3/2}}dx \stackrel{u=\ln(x)}{=} \int_0^\infty \frac{ue^u\cos(e^{2u})}{(e^u+1)^{3/2}}dx$$ Now, since $-1 \le \cos(x) \le 1$, simply note that $$ -\color{red}{\int_0^\infty \frac{ue^u}{(e^u+1)^{3/2}}du}\le\int_0^\infty \frac{ue^u\cos(e^{2u})}{(e^u+1)^{3/2}}du \le \color{red}{\int_0^\infty \frac{ue^u}{(e^u+1)^{3/2}}du}$$ And it is clear that $$ \sqrt{2}=\int_0^\infty \frac{e^u}{(e^u+1)^{3/2}}du\le\color{red}{\int_0^\infty \frac{ue^u}{(e^u+1)^{3/2}}du} \le \int_0^\infty \frac{ue^u}{(e^u+1)^2}du = \ln(2)$$

We therefore conclude that our original integral is convergent

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  • $\begingroup$ @JackD'Aurizio thanks, and good catch with the $dx$ typos. Fixed. Also, to avoid the inevitable question from the OP: Make a substitution and let $x = e^u+1$ on the left hand integral. The right hand integral is a bit tougher, but for a beginner I recommend integration by parts. Note that $u - \frac{u}{e^u+1} - \ln(e^u+1)$ is a primitive, which should give you a hint $\endgroup$ – Brevan Ellefsen Jun 21 '17 at 21:18
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    $\begingroup$ Isn't this a bit easier than we're making it. The integral $\int_e^\infty \frac{\log(x)}{(x+1)^{3/2}}\,dx$ converges since $|\log(x)|\le \frac{x^\alpha}{\alpha}$ for all $\alpha>0$. Choose any $\alpha<1/2$ and the integral converges using the comparison test. Therefore, the integral of interest is absolutely convergent. $\endgroup$ – Mark Viola Jun 21 '17 at 21:21
  • $\begingroup$ @MarkViola Good point; write up an answer using that idea and I will gladly upvote! $\endgroup$ – Brevan Ellefsen Jun 21 '17 at 21:22
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Note that $|\log(x)|<\frac{x^\alpha}{\alpha}$ for all $\alpha>0$. Hence, we see that

$$\left|\frac{\log(x)\cos(x^2)}{(x+1)^{3/2}}\right|\le \frac{1}{\alpha x^{3/2-\alpha}}$$

In particular, if we take $0<\alpha<1/2$, then we see by comparison with $\int_e^\infty \frac{1}{\alpha x^{3/2-\alpha}}\,dx$ that the integral $\int_e^\infty \frac{\log(x)\cos(x^2)}{(x+1)^{3/2}}$ is absolutely convergent.

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