Prove that $\int_{0}^{\infty}\frac{ \ln(x)\cos(\ln(x))}{x^2+1}dx=0$

Question

Prove that $$I=\int_{0}^{\infty}\frac{ \ln(x)\cos(\ln(x))}{x^2+1}dx=0$$

I used the substituion $x=e^t$ and the integral became $$I=\int_{-\infty}^{\infty}\frac{t\cos(t)}{e^{2t}+1}\cdot e^t\:\:dt$$ or $$I=\int_{-\infty}^{\infty}\frac{t\cos(t)}{e^{t}+e^{-t}}dt$$ or $$I=\int_{-\infty}^{\infty}\frac{t\cos(t)}{2\cosh(t)}dt$$

After this I tried different approaches but all in vain.

Any help is greatly appreciated.

Where does that problem come from? Why do you think that the integral is zero? — Martin R, Feb 02 '25 at 13:32
$I=\int \frac{t\cos t}{e^{2t}+1}e^t dt=\int \frac{t\cos t}{(e^{2t}+1)e^{-t}}dt=\int\frac{t\cos t}{(e^{t}+e^{-t})}dt=\int \frac{t\cos t}{2 \cosh t} dt$ — Sine of the Time, Feb 02 '25 at 13:34
I think this is obvious, if you let $x\mapsto 1/x$, you will have $I=-I$. — Nanayajitzuki, Feb 02 '25 at 13:35
This question is similar to: Why is $\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x =0$?. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Anne Bauval, Feb 02 '25 at 17:18

score 6 · Accepted Answer · answered Feb 02 '25 at 13:34

6

Actually,$$I=\int_{-\infty}^\infty\frac{t\cos(t)\color{red}{e^t}}{e^{2t}+1}\,\mathrm dt=\int_{-\infty}^\infty\frac{t\cos(t)}{e^t+e^{-t}}\,\mathrm dt.$$Since you are integrating an odd function, $I$ is indeed $0$.

answered Feb 02 '25 at 13:34

José Carlos Santos

440,053

i am so sorry..so silly of me..i forgot that e^t – Blue Cat Blues Feb 02 '25 at 13:35

Prove that $\int_{0}^{\infty}\frac{ \ln(x)\cos(\ln(x))}{x^2+1}dx=0$

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