Challenging problem: Find $a$ where $\int_0^\infty \frac{\cos(ax)\ln(1+x^2)}{\sqrt{1+x^2}}dx=0$.

Consider $$\underbrace{\int _0^{\infty }\frac{\cos \left(ax\right)}{\left(1+x^2\right)^b}\:dx}_{x=\frac{t}{a}}=a^{2b-1}\int _0^{\infty }\frac{\cos \left(t\right)}{\left(a^2+t^2\right)^b}\:dt$$ Now use the following identity that can be found here. $$K_v\left(z\right)=\frac{\Gamma \left(v+\frac{1}{2}\right)\left(2z\right)^v}{\sqrt{\pi }}\int _0^{\infty \:}\frac{\cos \left(t\right)}{\left(z^2+t^2\right)^{v+\frac{1}{2}}} dt$$ This leads to $$\int _0^{\infty }\frac{\cos \left(ax\right)}{\left(1+x^2\right)^b}\:dx=a^{2b-1}K_{b-\frac{1}{2}}\left(a\right)\frac{\sqrt{\pi }}{\Gamma \left(b\right)\left(2a\right)^{b-\frac{1}{2}}}$$ This means that $$\int _0^{\infty }\frac{\cos \left(ax\right)\ln \left(1+x^2\right)}{\sqrt{1+x^2}}\:dx=-\lim _{b\to \frac{1}{2}}\frac{\partial }{\partial b}a^{2b-1}K_{b-\frac{1}{2}}\left(a\right)\frac{\sqrt{\pi }}{\Gamma \left(b\right)\left(2a\right)^{b-\frac{1}{2}}}$$ Using mathematica to complete the calculations we are left with $$K_0\left(a\right)\left(-\ln \left(a\right)+\ln \left(2\right)+\psi \left(\frac{1}{2}\right)\right)-K^{\left(1,0\right)}_0\left(a\right)$$

Now you can check here that $$K^{\left(1,0\right)}_0\left(a\right)=0$$ Proof provided below.

Meaning overall $$=K_0\left(a\right)\left(-\ln \left(a\right)+\ln \left(2\right)-\gamma -2\ln \left(2\right)\right)$$ $$\boxed{\int _0^{\infty }\frac{\cos \left(ax\right)\ln \left(1+x^2\right)}{\sqrt{1+x^2}}\:dx=-K_0\left(a\right)\left(\ln \left(a\right)+\gamma +\ln \left(2\right)\right)}$$ Which agrees with the results proposed above.

Now answering the main point, $$-K_0\left(a\right)\left(\ln \left(a\right)+\gamma +\ln \left(2\right)\right)=0$$ $$\ln \left(2a\right)+\gamma =0$$ $$2a=e^{-\gamma }$$

We find that $\displaystyle a=\frac{e^{-\gamma}}{2}$

And so by plugging it in we can immediately see $$\int _0^{\infty }\frac{\cos \left(\frac{e^{-\gamma }}{2}x\right)\ln \left(1+x^2\right)}{\sqrt{1+x^2}}\:dx=-K_0\left(\frac{e^{-\gamma }}{2}\right)\left(-\gamma -\ln \left(2\right)+\gamma +\ln \left(2\right)\right)=0$$

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Proof of tools used.

$$K^{\left(1,0\right)}_0\left(a\right)=0$$

$$K_v\left(a\right)=\int _0^{\infty }e^{-a\cosh \left(t\right)}\cosh \left(vt\right)\:dt$$ Differentiating with respect to $v$ gives us $$K_v^{\left(1,0\right)}\left(a\right)=\int _0^{\infty }te^{-a\cosh \left(t\right)}\sinh \left(vt\right)\:dt$$ Now let $v=0$ $$K_0^{\left(1,0\right)}\left(a\right)=\int _0^{\infty }te^{-a\cosh \left(t\right)}\sinh \left(0\right)\:dt=0$$

$\displaystyle K_v\left(a\right)=\frac{\Gamma \left(v+\frac{1}{2}\right)\left(2a\right)^v}{\sqrt{\pi }}\int _0^{\infty }\frac{\cos \left(t\right)}{\left(a^2+t^2\right)^{v+\frac{1}{2}}}\:dt=\int _0^{\infty }e^{-a\cosh \left(t\right)}\cosh \left(vt\right)\:dt$

First consider $$I\left(a\right)=\int _0^{\infty }\frac{\cos \left(ax\right)}{\left(1+x^2\right)^v}\:dx$$

Well make use of the following gamma function representation $$\Gamma(v)={\left(1+x^{2}\right)}^{v}\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{v-1}du$$ Multiply $I\left(a\right)$ by $\Gamma(v)$ $$\Gamma(v)I(a)=\int_{0}^{\infty}\cos(ax)\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{v-1}dudx$$ $$=\int_{0}^{\infty}u^{v-1}e^{-u}\int_{0}^{\infty}e^{-x^{2}u}\cos(ax)dxdu=\frac{1}{2}\sqrt{{\pi}}\underbrace{\int_{0}^{\infty}u^{v-\frac{2}{2}}e^{-u-\frac{a^{2}}{4u}}du}_{u=\left(\frac{a}{2}\right)e^t}$$ $$=\frac{\sqrt{\pi}}{2}\frac{1}{\Gamma(v)}{\left(\frac{a}{2}\right)}^{v-\frac{1}{2}}\int_{-\infty}^{\infty}e^{-a\cosh(t)}e^{\left(v-\frac{1}{2}\right)t} dt$$ $$=\frac{\sqrt{\pi}}{\Gamma(v)}{\left(\frac{a}{2}\right)}^{v-\frac{1}{2}}\int_{0}^{\infty}e^{-a\cosh(t)}\cosh{\left(\left(v-\frac{1}{2}\right)t\right)} dt$$ $$\frac{\Gamma \left(v\right)}{\sqrt{\pi }}\:\left(\frac{2}{a}\right)^{v-\frac{1}{2}}\int _0^{\infty }\frac{\cos \left(ax\right)}{\left(1+x^2\right)^v}\:dx=\int_{0}^{\infty}e^{-a\cosh(t)}\cosh{\left(\left(v-\frac{1}{2}\right)t\right)} dt$$ $$\frac{\Gamma \left(v+\frac{1}{2}\right)}{\sqrt{\pi }}\:\left(\frac{2}{a}\right)^v\int _0^{\infty }\frac{\cos \left(ax\right)}{\left(1+x^2\right)^{v+\frac{1}{2}}}\:dx=\int _0^{\infty }e^{-a\cosh \left(t\right)}\cosh \left(vt\right)\:dt$$

The integral equals $$ K_0(a) (\gamma+\log(2)+\log(a)) \tag{*} $$ (where $K_0(a)$ is a modified Bessel function, assume $a>0$, $a<0$ follows by symmetry), which can be shown by integration under the integral sign together with $K_0'(a)=-K_1(a)$.

Since Bessel K's have no zeros, we can equate the bracket in (*) to zero and get

$$ a=\pm\frac{e^{-\gamma}}{2}\approx\pm 0.28073\,, $$

which is the same as numercis suggests (see comments to question).

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

$$\ln|2qe^{\gamma}|=0\Longrightarrow q=\pm\frac12e^{-\gamma}.$$