TITLE: Challenging problem: Find $a$ where $\int_0^\infty \frac{\cos(ax)\ln(1+x^2)}{\sqrt{1+x^2}}dx=0$.
QUESTION [10 upvotes]: What is the value of $a\in\mathbb{R}$ that makes the following integral true
$$\int_0^\infty \frac{\cos(ax)\ln(1+x^2)}{\sqrt{1+x^2}}dx=0\,?$$

This question was proposed by my friend Khalef Ruhemi and I have no idea how to approach it but all I tried is setting $x=\tan\theta$ and I don't know how to continue after that. Also I noticed that the integrand is an even function and again I don't know how to make use of this fact. Any help would be much appreciated.

REPLY [2 votes]: Solution due to Khalef Ruhemi without using any kind of software:
Define
$$f(p,q)=\int_0^\infty\frac{\cos(qx)}{(1+x^2)^p}dx,\quad p>0, q\ne0$$
By $$\frac{1}{(1+x^2)^p}=\frac{1}{\Gamma(p)}\int_0^\infty y^{p-1}e^{-(1+x^2)y}dy$$
We  have
$$f(p,q)=\frac{1}{\Gamma(p)}\int_0^\infty y^{p-1} e^{-y}\underbrace{\left(\int_0^\infty e^{-x^2y}\cos(qx) dx\right)}_{x^2y=t^2}dy$$
$$=\frac{1}{\Gamma(p)}\int_0^\infty y^{p-\frac32} e^{-y}\left(\int_0^\infty e^{-t^2}\cos\left(\frac{qt}{\sqrt{y}}\right)dt\right)dy$$
$$=\frac{\sqrt{\pi}}{2\Gamma(p)}\int_0^\infty y^{p-\frac32} e^{-(y+\frac{q^2}{4y})}dy\tag1$$
$$\overset{\frac{q^2}{4y}=x}{=}\frac{\sqrt{\pi}}{2\Gamma(p)}\left|\frac{q}{2}\right|^{2p-1}\int_0^\infty x^{-p-\frac12}e^{-(x+\frac{q^2}{4x})}dx$$
$$=\frac{\Gamma(1-q)}{\Gamma(p)}\left|\frac{q}{2}\right|^{2p-1}\underbrace{\left(\frac{\sqrt{\pi}}{2\Gamma(1-p)}\int_0^\infty x^{-p-\frac12}e^{-(x+\frac{q^2}{4x})}dx\right)}_{=f(1-p,q)\ \text{by} (1)}$$
Thus,
$$f(p,q)=\frac{\Gamma(1-q)}{\Gamma(p)}\left|\frac{q}{2}\right|^{2p-1}f(1-p,q)$$
or,
$$\int_0^\infty\frac{\cos(qx)}{(1+x^2)^p}dx=\frac{\Gamma(1-q)}{\Gamma(p)}\left|\frac{q}{2}\right|^{2p-1}\int_0^\infty\frac{\cos(qx)}{(1+x^2)^{1-p}}dx,\quad 0<p<1\tag2$$
Note that $0<p<1$ follows from the fact that $p>0$ and $1-p>0$.
Next, differentiate both sides of $(2)$ with respect to $p$ then let $p\to 1/2$ we have
$$\int_0^\infty\frac{\cos(qx)\ln(1+x^2)}{\sqrt{1+x^2}}dx=-\ln|2qe^{\gamma}|\int_0^\infty\frac{\cos(qx)}{\sqrt{1+x^2}}dx$$
Finally, since the LHS integral is equal to zero, we have
$$\ln|2qe^{\gamma}|=0\Longrightarrow q=\pm\frac12e^{-\gamma}.$$