Show $\int_0^{\pi/2}\int_0^1 e^{t+t^{\tan\theta}}dtd\theta=\frac{\pi}{4}(e^2-1)$

Question

A friend gave me this double integral a while ago, and I couldn't figure out how to solve it. $$\int_0^{\pi/2} \int_0^1 e^{t+t^{\tan\theta}}\,dt d\theta=\dfrac{\pi}{4}\left(e^2-1\right)$$ I tried using Fubini's theorem, but I did not know what to do from there $$\int_0^1e^t\int_0^{\pi/2} e^{t^{\tan\theta}}\,d\theta dt=\int_0^1 e^t\int_0^{\pi/2}\sum_{n=0}^\infty \frac{t^{n\tan\theta}}{n!}\,d\theta dt $$ $$\int_0^{\pi/2}a^{\tan\theta}\,d\theta=?$$ Wolfram Alpha gives a closed form in terms of exponential integral functions, but it seems challenging to solve the sum at the end involving exponential integral functions. Please help

Answer 1

First begin with the substitution $s=\tan\theta\implies \frac{1}{1+s^2}ds=d\theta$, giving us \begin{align*}I&:=\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}e^{t+t^{\tan\theta}}\, dt\, d\theta \\ &\,=\int_{0}^{\infty}\int_{0}^{1}\frac{\exp(t+t^s)}{1+s^2}\, dt\, ds\stackrel{s\to 1/s}{=}\int_{0}^{\infty}\int_{0}^{1}\frac{\exp(t+t^{1/s})}{1+s^2}\, dt\, ds \\ &\!\!\stackrel{t\to t^s}{=}\int_{0}^{\infty}\int_{0}^{1}\frac{\exp(t+t^s)s t^{s-1}}{1+s^2}\, dt\, ds\\ &=\int_{0}^{\infty}\int_{0}^{1}\frac{\exp(t+t^s)(s t^{s-1}+1)}{1+s^2}\, dt\, ds-\int_{0}^{\infty}\int_{0}^{1}\frac{\exp(t+t^s)}{1+s^2}\, dt\, ds \end{align*} Now notice that since $\frac{d}{dt}\exp(t+t^s)=\exp(t+t^s)(st^{s-1}+1)$, then integrating with respect to $t$ on the first integral is trivial, and gives us $$I=\int_{0}^{\infty}\frac{e^2-1}{1+s^2}\, ds-I\implies I=\frac{\pi}{4}(e^2-1),$$ as desired. $\square$