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

Solution due to Khalef Ruhemi without using any kind of software:

Define

$$f(p,q)=int_0^inftyfrac{cos(qx)}{(1+x^2)^p}dx,quad p>0, qne0$$

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

$$f(p,q)=frac{1}{Gamma(p)}int_0^infty y^{p-1} e^{-y}underbrace{left(int_0^infty e^{-x^2y}cos(qx) dxright)}_{x^2y=t^2}dy$$

$$=frac{1}{Gamma(p)}int_0^infty y^{p-frac32} e^{-y}left(int_0^infty e^{-t^2}cosleft(frac{qt}{sqrt{y}}right)dtright)dy$$

$$=frac{sqrt{pi}}{2Gamma(p)}int_0^infty y^{p-frac32} e^{-(y+frac{q^2}{4y})}dytag1$$

$$overset{frac{q^2}{4y}=x}{=}frac{sqrt{pi}}{2Gamma(p)}left|frac{q}{2}right|^{2p-1}int_0^infty x^{-p-frac12}e^{-(x+frac{q^2}{4x})}dx$$

$$=frac{Gamma(1-q)}{Gamma(p)}left|frac{q}{2}right|^{2p-1}underbrace{left(frac{sqrt{pi}}{2Gamma(1-p)}int_0^infty x^{-p-frac12}e^{-(x+frac{q^2}{4x})}dxright)}_{=f(1-p,q) text{by} (1)}$$

Thus,

$$f(p,q)=frac{Gamma(1-q)}{Gamma(p)}left|frac{q}{2}right|^{2p-1}f(1-p,q)$$

or,

$$int_0^inftyfrac{cos(qx)}{(1+x^2)^p}dx=frac{Gamma(1-q)}{Gamma(p)}left|frac{q}{2}right|^{2p-1}int_0^inftyfrac{cos(qx)}{(1+x^2)^{1-p}}dx,quad 0<p<1tag2$$

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 $pto 1/2$ we have

$$int_0^inftyfrac{cos(qx)ln(1+x^2)}{sqrt{1+x^2}}dx=-ln|2qe^{gamma}|int_0^inftyfrac{cos(qx)}{sqrt{1+x^2}}dx$$

Finally, since the LHS integral is equal to zero, we have

$$ln|2qe^{gamma}|=0Longrightarrow q=pmfrac12e^{-gamma}.$$

Consider $$underbrace{int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^b}:dx}_{x=frac{t}{a}}=a^{2b-1}int _0^{infty }frac{cos left(tright)}{left(a^2+t^2right)^b}:dt$$ Now use the following identity that can be found here. $$K_vleft(zright)=frac{Gamma left(v+frac{1}{2}right)left(2zright)^v}{sqrt{pi }}int _0^{infty :}frac{cos left(tright)}{left(z^2+t^2right)^{v+frac{1}{2}}} dt$$ This leads to $$int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^b}:dx=a^{2b-1}K_{b-frac{1}{2}}left(aright)frac{sqrt{pi }}{Gamma left(bright)left(2aright)^{b-frac{1}{2}}}$$ This means that $$int _0^{infty }frac{cos left(axright)ln left(1+x^2right)}{sqrt{1+x^2}}:dx=-lim _{bto frac{1}{2}}frac{partial }{partial b}a^{2b-1}K_{b-frac{1}{2}}left(aright)frac{sqrt{pi }}{Gamma left(bright)left(2aright)^{b-frac{1}{2}}}$$ Using mathematica to complete the calculations we are left with $$K_0left(aright)left(-ln left(aright)+ln left(2right)+psi left(frac{1}{2}right)right)-K^{left(1,0right)}_0left(aright)$$

Now you can check here that $$K^{left(1,0right)}_0left(aright)=0$$ Proof provided below.

Meaning overall $$=K_0left(aright)left(-ln left(aright)+ln left(2right)-gamma -2ln left(2right)right)$$ $$boxed{int _0^{infty }frac{cos left(axright)ln left(1+x^2right)}{sqrt{1+x^2}}:dx=-K_0left(aright)left(ln left(aright)+gamma +ln left(2right)right)}$$ Which agrees with the results proposed above.

Now answering the main point, $$-K_0left(aright)left(ln left(aright)+gamma +ln left(2right)right)=0$$ $$ln left(2aright)+gamma =0$$ $$2a=e^{-gamma }$$

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

And so by plugging it in we can immediately see $$int _0^{infty }frac{cos left(frac{e^{-gamma }}{2}xright)ln left(1+x^2right)}{sqrt{1+x^2}}:dx=-K_0left(frac{e^{-gamma }}{2}right)left(-gamma -ln left(2right)+gamma +ln left(2right)right)=0$$

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Proof of tools used.

$$K^{left(1,0right)}_0left(aright)=0$$

$$K_vleft(aright)=int _0^{infty }e^{-acosh left(tright)}cosh left(vtright):dt$$ Differentiating with respect to $v$ gives us $$K_v^{left(1,0right)}left(aright)=int _0^{infty }te^{-acosh left(tright)}sinh left(vtright):dt$$ Now let $v=0$ $$K_0^{left(1,0right)}left(aright)=int _0^{infty }te^{-acosh left(tright)}sinh left(0right):dt=0$$

$displaystyle K_vleft(aright)=frac{Gamma left(v+frac{1}{2}right)left(2aright)^v}{sqrt{pi }}int _0^{infty }frac{cos left(tright)}{left(a^2+t^2right)^{v+frac{1}{2}}}:dt=int _0^{infty }e^{-acosh left(tright)}cosh left(vtright):dt$

First consider $$Ileft(aright)=int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^v}:dx$$

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 $Ileft(aright)$ by $Gamma(v)$ $$Gamma(v)I(a)=int_{0}^{infty}cos(ax)int_{0}^{infty}e^{-left(1+x^{2}right)u} u^{v-1}dudx$$ $$=int_{0}^{infty}u^{v-1}e^{-u}int_{0}^{infty}e^{-x^{2}u}cos(ax)dxdu=frac{1}{2}sqrt{{pi}}underbrace{int_{0}^{infty}u^{v-frac{2}{2}}e^{-u-frac{a^{2}}{4u}}du}_{u=left(frac{a}{2}right)e^t}$$ $$=frac{sqrt{pi}}{2}frac{1}{Gamma(v)}{left(frac{a}{2}right)}^{v-frac{1}{2}}int_{-infty}^{infty}e^{-acosh(t)}e^{left(v-frac{1}{2}right)t} dt$$ $$=frac{sqrt{pi}}{Gamma(v)}{left(frac{a}{2}right)}^{v-frac{1}{2}}int_{0}^{infty}e^{-acosh(t)}cosh{left(left(v-frac{1}{2}right)tright)} dt$$ $$frac{Gamma left(vright)}{sqrt{pi }}:left(frac{2}{a}right)^{v-frac{1}{2}}int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^v}:dx=int_{0}^{infty}e^{-acosh(t)}cosh{left(left(v-frac{1}{2}right)tright)} dt$$ $$frac{Gamma left(v+frac{1}{2}right)}{sqrt{pi }}:left(frac{2}{a}right)^vint _0^{infty }frac{cos left(axright)}{left(1+x^2right)^{v+frac{1}{2}}}:dx=int _0^{infty }e^{-acosh left(tright)}cosh left(vtright):dt$$

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

$$ a=pmfrac{e^{-gamma}}{2}approxpm 0.28073,, $$

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