Dtyjlui

Does anyone know how to solve this improper integral?

$int_{0}^{infty} frac{a^{k^{2}}cos(bk)}{c+k^{2}}dk$, where $ain (0,1)$, $b in mathbb{R}$, and $c geqslant 0$.

Thank you!

If you are familiar with Fourier transforms you may try this:

• $I:=int_{0}^{infty} frac{a^{x^{2}}cos(bx)}{c+x^{2}}dx=$ {by parity} =$frac{1}{2} int_{mathbb{R}} frac{a^{x^{2}}cos(bx)}{c+x^{2}}dx$. Rewrite $a^{x^2}=e^{x^2 ln a}=e^{-qx^2}$, where $q>0$, since $ain (0,1)$.




• Remember $cos(bx)=frac{e^{ibx}+e^{-ibx}}{2}$, so $2I=frac{1}{2} int_{mathbb{R}} frac{e^{-qx^2}e^{-i(-b)x}}{c+x^{2}}dx+frac{1}{2} int_{mathbb{R}} frac{e^{-qx^2}e^{-ibx}}{c+x^{2}}dx$. Calling $f(x)=frac{e^{-qx^2}}{c+x^{2}} in L^1(mathbb{R})$, there exists its Fourier transform $mathcal{F} { f}$ and $2I=frac{1}{2}{ mathcal{F} { f}(-b)+mathcal{F} { f}(b)}$.




• Since $f$ is even $mathcal{F} { f}$ is even, so $2I=mathcal{F} { f}(-b)$




• Now you need to compute $mathcal{F} { f}$, for example solving the relative complex integral using the residue theorem on a semicircumference, and the appropriate case of Jordan's lemma: the integral should turn out to be $2I=frac{pi e^{-bsqrt c + qc}}{sqrt c}=frac{pi e^{-b sqrt c}}{a^csqrt c}$ (I assumed here $b>0$, the other case is analogous)

This is not a complete answer but it still may be useful to you.

Let $c=alpha^2$. Considering the symmetry of the integrand, using the fact that $frac{1}{alpha^2+k^2}=int_0^infty e^{-nu(alpha^2+k^2)},dnu$ and reversing the order of integration, we end up with the double integral
$$frac{1}{2}int_0^infty dnu ,e^{-alpha^2nu} int_{-infty}^infty dk ,e^{-k^2(nu-log a)}cos(bk)$$
Note that we can replace $cos(bk)$ by $e^{-ibk}$ in the integral as the imaginary part will cancel because it will be an odd fonction integrated over an even interval. After doing so, letting $nu-log a = p$, and completing the square in the exponent, we are left with
$$frac{1}{2}int_0^infty dnu ,e^{-alpha^2 nu} e^{-b^2/4p}int_{-infty}^infty dk,e^{-p(k-ibk/2p)^2}$$
Choosing a simple rectangular contour in the complex plane and applying Cauchy's theorem will show that $int_{mathbb{R}}e^{-p(k-ibk/p)^2}=int_{mathbb{R}}e^{-pk^2}$. Using this last equality as well as the standard $int_{mathbb{R}}dk, e^{-pk^2}=sqrt{frac{pi}{p}}$, we simply need to compute the single integral
$$sqrt{frac{pi}{4}} int_0^infty frac{dnu}{sqrt{nu-log a}}, e^{-alpha^2nu-b^2/4(nu-log a)}$$

I'm not sure if this has simplified the integral, but it's one attempt at it. The exponent makes me think of Glasser's master theorem