Methods to evaluate $\int_{0}^{\infty}{e^{-\sigma x^n}\cos(ax^n)\, dx}$

Question

I was trying some things out to evaluate $\int_{0}^{\infty}{e^{-\sigma x^n}\cos(ax^n)\, dx}$ but I couldn't figure out some of the details. For example :

$$\begin{align}I&=\int_{0}^{\infty}{e^{-\sigma x^n}\cos(ax^n)\, dx}\\ &=\Re\int_{0}^{\infty}{e^{-x^n(\sigma-ai)}\, dx}\\ &=\Re\frac{1}{n}\int_{0}^{\infty}{e^{-t(\sigma-ia)}t^{1/n-1}\, dt}\\ &=\Re\frac{1}{(\sigma - ia)^{n+1}}\Gamma\left(\frac{1}{n} +1\right)\end{align}$$

with the last bit using $\int_{0}^{\infty}{e^{-st}t^n\,dt}=s^{-(n+1)}\Gamma(n+1)$ and $n\Gamma(n)=\Gamma(n+1)$. The formula doesn't seem right, but WFA says $\int_{0}^{\infty}e^{-5x^3}\cos(3x^3)\, dx \approx0.469$ but the formula says it's $-0.000480\dots$ (checked [here][1])

Another attempt :

$$ \begin{align} I&=\int_{0}^{\infty}{e^{-\sigma x^n}\cos(ax^n)\, dx}\\ &=\int_{0}^{\infty}{e^{-\sigma x^n}\sum_{k=0}^{\infty}{(-1)^k}\frac{(ax^n)^{2k}}{(2k)!}\, dx}\\ &=\sum_{k=0}^{\infty}{(-1)^k\frac{a^{2k}}{(2k)!}\int_{0}^{\infty}e^{-\sigma x^n}(x^n)^{2k}\,dx}\\ &=\frac{1}{n}\sum_{k=0}^{\infty}{(-1)^k\frac{a^{2k}}{(2k)!}}\int_{0}^{\infty}{e^{-\sigma t}t^{2k+1/n-1}\, dt}\\ &=\frac{1}{n\sigma}\sum_{k=0}^{\infty}{(-1)^k\frac{a^{2k}}{(2k)!}}\int_{0}^{\infty}{e^{-u}\left(\frac{u}{\sigma}\right)^{2k+1/n-1}\, du}\\ &=\frac{1}{n\sigma^{1/n}}\sum_{k=0}^{\infty}{(-1)^k\frac{\left(\frac{a}{\sigma}\right)^{2k}}{(2k)!}\Gamma\left(2k+\frac{1}{n}\right)} \end{align} $$

Can the sum be simplified in this case and how can the first method be amended?

EDIT : Found the issue with the first attempt, instead I should've used $\int_{0}^{\infty}{e^{-st}t^{1/n-1}\, dt}=s^{-1/n}\Gamma(1/n)$ which then yields the correct answer. So $I=\Re\frac{1}{(\sigma - ia)^{1/n}}\Gamma\left(\frac{1}{n} +1\right)$ Maybe it can be simplified further? [1]: https://www.wolframalpha.com/input?i=int%20e%5E%28-5x%5E3%29cos%284x%5E3%29%200..inf

Answer 1

Using $\cos t= \Re (e^{-ti})$, we have $$ \begin{aligned} I & =\int_0^{\infty} e^{-\sigma x^n} \cos \left(a x^n\right) d x \\ & =\Re \int_0^{\infty} e^{-\sigma x^n} e^{i a x^n} d x \\ & =\Re \int_0^{\infty} e^{-(\sigma-i a) x^n} d x \end{aligned} $$ Letting $y= x^n$ converts the integral $I$ into a Gamma function. $$ \begin{aligned} I & =\Re\int_0^{\infty} e^{-(\sigma-i a) y} \cdot \frac{1}{n} y^{\frac{1}{n}-1} d y \\ & =\frac{1}{n} \Re \int_0^{\infty} y^{\frac{1}{n}-1} e^{-(\sigma-i a) y} d y \\ &=\frac{\Gamma\left(\frac{1}{n}+1\right)}{n} \Re\left[\frac{1}{(\sigma-i a)^{\frac{1}{n}}}\right] \end{aligned} $$ Expressing the last reciprocal into its polar form as $$ \begin{aligned} & \frac{1}{(\sigma-i a)^{\frac{1}{n}}}=\frac{1}{\left(\sqrt{\sigma^2+a^2} e^{-i \tan ^{-1} \frac{a}{\sigma}}\right)^{\frac{1}{n}}} =\frac{e^{\frac{i}{n} \tan ^{-1} \frac{a}{\sigma}}}{\left(\sigma^2+a^2\right)^{\frac{1}{2n}}} \end{aligned} $$ Now we can conclude that $$ I=\frac{\Gamma\left(\frac{1}{n}+1\right)}{n} \Re\left[\frac{1}{(\sigma-i a)^{\frac{1}{n}}}\right]=\frac{\Gamma\left(\frac{1}{n}+1\right)\cos \left(\frac{1}{n} \tan ^{-1} \frac{a}{\sigma}\right)}{n\left(\sigma^2+a^2\right)^{\frac{1}{2 n}}} $$

Answer 2

Hint:

The change of variables $x=u^{1/n}$, $dx=\frac{1}{n}u^{\frac1n-1}\,du$ yields \begin{align} \frac{1}{n}\int^\infty_0e^{-\sigma u}u^{\frac1n-1}\cos(au)du=\frac{\Gamma(\frac1n)}{n\sigma^{\frac1n}}\int^\infty_0 \cos(au)f(u;\frac1n,\sigma)\,du\tag{0}\label{zero} \end{align} where $f(u;\alpha,\beta)=\frac{\beta^\alpha}{\Gamma(\alpha)}e^{-\beta u}u^{\alpha-1}$ is the densitty function of the $\operatorname{Gamma}(\alpha,\beta)$ distribution. Here $\alpha=\frac1n$ and $\beta=\sigma$.

The integral in the right-hand-side of \eqref{zero} is the real part of the characteristic function (Fourier transform) of the $\operatorname{Gamma}(\alpha,\beta)$ distribution:

$$\phi(a;\alpha,\beta)=\int^\infty_0 e^{iau}f(u;\alpha,\beta)\,du= {\displaystyle \left(1-{\frac {ia}{\beta }}\right)^{-\alpha }}$$

The latter can be obtained by computing first the Laplace transform of the gamma distribution $$ L(a)=\int^\infty_0 e^{-au}f(u;\alpha,\beta)\,du= \left(1+{\frac {a}{\beta }}\right)^{-\alpha }$$ and then, using analytic continuation to get $$\phi(a)=L(-ia)$$

Th obtain the real part of $\Big(1-i\frac{a}{\beta}\Big)^{-\alpha}$ we take the principal branch of logarithm and notice that $$ \Big(1-i\frac{a}{\beta}\Big)^{-\alpha}=\exp\Big(-\alpha\log\big(1-i\frac{a}{\beta}\big)\Big)$$ and $$\log\big(1-i\frac{a}{\beta}\big)=\log\big|1-i\frac{a}{\beta}\big|+i\arg\big(1-i\frac{a}{\beta}\big)$$ which is easy to estimate.