Infinite integral of $e^{-\alpha t^\beta} \cos(\omega t)$

Question

I am attempting to calculate the integral \begin{equation} I=\int_0^{\infty} e^{-\alpha t^\beta} \cos(\omega t) dt, \end{equation} where $\alpha,\beta >0$, and $\omega$ are just parameters.

My idea is to employ the formula $\int_0^{\infty} x^{\nu-1} e^{-\mu x^p} d x=\frac{1}{p} \mu^{-\frac{\nu}{p}} \Gamma\left(\frac{\nu}{p}\right) \quad[\operatorname{Re} \mu>0, \quad \operatorname{Re} \nu>0, \quad p>0]$ (see e.g., Table of Integrals, Series, and Products 7th Edition, p370), and the Taylor series of the cosine function.

Then, we have \begin{aligned} I & =\int_0^{\infty} e^{-\alpha t^\beta} \cos(\omega t) dt \\ & =\sum_{k=0}^{\infty} (-1)^k \frac{\omega^{2k}}{(2k)!}\int_0^{\infty} t^{2k}e^{-\alpha t^\beta} dt \\ & =\frac{1}{\alpha^{1/\beta} \beta}\sum_{k=0}^{\infty} (-1)^k \frac{\omega^{2k}}{(2k)!} \alpha^{-\frac{2k}{\beta}} \Gamma \left(\frac{2k+1}{\beta}\right) \\ & = ? \end{aligned}

However, I am not sure if such an infinite series can be simplified or not.

I have tried other regular methods, such as the residue theorem and Laplace transform, but also have no ideas to get an analytical result of the integral $I$.

Answer 1

$$I=\int_0^\infty e^{-\alpha t^{\beta }} \cos ( \omega t )\,dt$$

Let $$x= \omega t \qquad \text{and} \qquad \gamma=\alpha \omega ^{-\beta }$$ $$I=\frac 1 \omega\int_0^\infty e^{-\gamma x^{\beta }}\cos (x)\,dx=\frac 1 \omega\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!} \int_0^\infty e^{-\gamma x^{\beta }}\,x^{2n}\,dx$$ as you wrote $$I=\frac 1 {\beta\omega\gamma^{\frac 1 \beta}}\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}\,\,\gamma ^{-\frac{2 n}{\beta }}\,\,\Gamma \left(\frac{2 n+1}{\beta }\right)$$ which must have a closed form in terms of hypergeometric functions (may be, some of them can reduce to Kelvin or Airy functions).

I have not been able to extract the formula for the general case.

Just an example for $\alpha=2$, $\beta=3$ and $\omega=4$.

$$I=-\frac{4}{3} \, _1F_4\left(1;\frac{4}{6},\frac{5}{6},\frac{7}{6},\frac{8}{6};- \frac{64}{729}\right)+$$ $$\frac{2\pi}{9} \left(\text{bei}_{\frac{1}{3}}\left(\frac{8 \sqrt{\frac{2}{3}}}{3}\right)-\text{ber}_{\frac{1}{3}}\left(\frac {8 \sqrt{\frac{2}{3}}}{3}\right)\right)-$$ $$\frac{2\pi}{9} \left(\text{bei}_{-\frac{1}{3}}\left(\frac{8 \sqrt{\frac{2}{3}}}{3}\right)-\text{ber}_{-\frac{1}{3}}\left(\frac {8 \sqrt{\frac{2}{3}}}{3}\right)\right)$$

Answer 2

With help of Mathematica: $$\int_0^{\infty } \exp \left(-\alpha t^{\beta }\right) \cos (\omega t) \, dt=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_{\alpha }\left[\exp \left(-\alpha t^{\beta }\right) \cos (\omega t)\right](s) \, dt\right](\alpha )=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \left(t^{\beta }\right)^{-s} \cos (t \omega ) \Gamma (s) \, dt\right](\alpha )=\\\mathcal{M}_s^{-1}\left[| \omega | ^{-1+s \beta } \Gamma (s) \Gamma (1-s \beta ) \sin \left(\frac{\pi s \beta }{2}\right)\right](\alpha )=\\\mathcal{M}_s^{-1}\left[\frac{\pi | \omega | ^{-1+s \beta } \Gamma (s) \Gamma (1-s \beta )}{\Gamma \left(\frac{s \beta }{2}\right) \Gamma \left(1-\frac{s \beta }{2}\right)}\right](\alpha )=\\-\frac{\pi H_{1,3}^{1,1} \! \left(\left[\left[\left[0, \beta \right]\right], \left[\right]\right], \left[\left[\left[0, 1\right]\right], \left[\left[0, -\frac{\beta}{2}\right], \left[1, \frac{\beta}{2}\right]\right]\right],{\alpha | \omega |}^{-\beta}\right)}{{| \omega |}} $$ where $ H_{1,3}^{1,1}$ is FoxH function.

MMA code:

Integrate[ Exp[-\[Alpha] *t^\[Beta]]*Cos[\[Omega]*t], {t, 0, Infinity}] == -((\[Pi] FoxH[{{{0, \[Beta]}}, {}}, {{{0, 1}}, {{0, -(\[Beta]/2)}, {1, \[Beta]/ 2}}}, \[Alpha] Abs[\[Omega]]^-\[Beta]])/Abs[\[Omega]])