Integral involving a Gaussian with a trig function in the exponent

Question

I have

$$ \int e^{-\beta\sec^2\left(x\right)}dx\ $$

and tried the substitution $x=\tan^{-1}u$ , giving

$$ e^{-\beta}\int\frac{dx}{1+x^{2}}e^{-\beta x^{2}}\ $$

...is this easier?

Answer 1

$\int e^{-\beta\sec^2 x}~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\beta^n\sec^{2n}x}{n!}dx$

$=\int dx+\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n\beta^n\sec^{2n}x}{n!}dx$

$=\int dx+\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n\beta^n\sec^{2n-2}x}{n!}d(\tan x)$

$=\int dx+\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n\beta^n(1+\tan^2x)^{n-1}}{n!}d(\tan x)$

$=\int dx+\int\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\beta^nC_{k-1}^{n-1}\tan^{2k-2}x}{n!}d(\tan x)$

$=x+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\beta^n(n-1)!\tan^{2k-1}x}{n!(n-k)!(k-1)!(2k-1)}+C$

Answer 2

If your interval of integration is the entire real line, then you may evaluate this integral using Parseval/Plancherel's theorem, i.e.,

$$\int_{-\infty}^{\infty} dx \, f(x) g(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G(k)$$

where $F$ and $G$ are the Fourier transforms of $f$ and $g$ respectively. In this case,

$$f(x) = \frac1{1+x^2} \implies F(k) = \pi \, e^{-|k|}$$ $$g(x) = e^{-\beta x^2} \implies G(k) = \sqrt{\frac{\pi}{\beta}} e^{-k^2/(4 \beta)}$$

So your integral becomes

$$\frac12 \sqrt{\frac{\pi}{\beta}} e^{-\beta} \int_{-\infty}^{\infty} dk \, e^{-k^2/(4 \beta)} e^{-|k|}$$

Split the integral up into two pieces and you will find the result is a sum of two error functions.