Closed form of :$\int\tan ( e^{-x²})\ d x$ Over reals

Question

I w'd surprised if this integral $\int\tan ( e^{-x²})\ d x$ has a closed form since :$\int_{-\infty}^{+\infty }\tan ( e^{-x²})\ d x$ is assumed as a constant by wolfram alpha and The inverse calculator didn't give me any thing such that it's value is $2.27591....$ , then any way to show that if it has a closed form , or it is just a constant ?

Answer 1

Knowing that:

$$\tan (x)=\sum _{n=1}^{\infty } \frac{\left((-1)^n \left(4^n-16^n\right) B_{2 n}\right) x^{2 n-1}}{\Gamma (1+2 n)} $$ so:

$$\int \tan \left(\exp \left(-x^2\right)\right) \, dx=\\\int \left(\sum _{n=1}^{\infty } \frac{(-1)^n \left(4^n-16^n\right) e^{-(-1+2 n) x^2} B_{2 n}}{\Gamma (1+2 n)}\right) \, dx=\\\sum _{n=1}^{\infty } \int \frac{(-1)^n \left(4^n-16^n\right) e^{-(-1+2 n) x^2} B_{2 n}}{\Gamma (1+2 n)} \, dx=\\\sum _{n=1}^{\infty } \frac{(-1)^{1+n} 2^{-1+2 n} \left(-1+4^n\right) \sqrt{\pi } B_{2 n} \text{erf}\left(\sqrt{-1+2 n} x\right)}{\sqrt{-1+2 n} \Gamma (1+2 n)}$$

and for define integral:

$$\sum _{n=1}^{\infty } \frac{(-1)^{1+n} 4^n \left(-1+4^n\right) \sqrt{\pi } B_{2 n}}{\sqrt{-1+2 n} \Gamma (1+2 n)}\approx 2.27591$$

where: $B_{2n}$ is Bernoulli numbers.

I doubt there's a closed form for the Sum.