In Mathematica,
$$\int_{0}^\infty \frac{\sin(ax)x}{(x^2+1)^c} dx
=\frac{2^{\frac{1}{2}-c}a^{-\frac{1}{2}+c}\pi^{\frac{1}{2}}\operatorname{BesselK}[-\frac{3}{2}+c,a])}{\Gamma[c]} ,$$
where a is a positive real number and $c>\frac{1}{2}.$
I want to prove this, but I can't.
If anyone knows the proof of the above definite integral,
Thank you for your instruction.
Best Answer
This is not a complete solution, but may be one reasonable approach to it.
Firts define: $$I(a)=\int_{0}^{\infty} \frac{\cos(ax)}{{\left(1+x^{2}\right)}^{c}}dx$$ Which can itself be written in terms of $K_{v}(x)$
Taking the derivative of $I(a)$ with respect to a gives us $$\frac{dI(a)}{da}=-\int_{0}^{\infty} \frac{x\sin(ax)}{{\left(1+x^{2}\right)}^{c}}dx$$
Now, consider the following integral, easily verified by a change of variable
$$\Gamma(c)={\left(1+x^{2}\right)}^{c}\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{c-1}du$$
now multiply $I(a)$ by $\Gamma(c)$
$$\Gamma(c)I(a)=\int_{0}^{\infty}cos(ax)\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{c-1}dudx$$
Swap the integrals and distribute the exponential
$$\Gamma(c)I(a)=\int_{0}^{\infty}u^{c-1}e^{-u}\int_{0}^{\infty}e^{-x^{2}u}\cos(ax)dxdu$$
The inner integral has the following solution
$$\frac{1}{2}\sqrt{\frac{\pi}{u}}e^{-\frac{a^{2}}{4u}}$$
Giving us:
$$\Gamma(c)I(a)=\frac{\sqrt{\pi}}{2}\int_{0}^{\infty}u^{c-\frac{2}{2}}e^{-u-\frac{a^{2}}{4u}}du$$
Make the following substitution $u=\left(\frac{x}{2}\right)e^{v}$
$$ I(a)=\frac{\sqrt{\pi}}{2}\frac{1}{\Gamma(c)}{\left(\frac{a}{2}\right)}^{c-\frac{1}{2}}\int_{-\infty}^{\infty}e^{-a\cosh(v)}e^{\left(c-\frac{1}{2}\right)v} dv$$
because $e^{\left(c-\frac{1}{2}\right)v} = \cosh\left(\left(c-\frac{1}{2}\right)v\right)+\sinh\left(\left(c-\frac{1}{2}\right)v\right)$, and since $\sinh\left(\left(c-\frac{1}{2}\right)v\right)$ is an odd function of $v$ gives us the final result:
$$\boxed{I(a)=\frac{\sqrt{\pi}}{\Gamma(c)}{\left(\frac{a}{2}\right)}^{c-\frac{1}{2}}\int_{0}^{\infty}e^{-a\cosh(v)}\cosh{\left(\left(c-\frac{1}{2}\right)v\right)} dv}$$
$$\boxed{I(a)=\frac{\sqrt{\pi}}{\Gamma(c)}{\left(\frac{a}{2}\right)}^{c-\frac{1}{2}}K_{c-\frac{1}{2}}(x)}$$
The last integral is an Integral representation 2 of $K_{v}(x)$
Now, take the derivative of this expression with respect to $a$.