Find $\int_0^{\frac{\pi}{4}}e^{\sec^2 x}dx$ How can we find 
$$\int_0^{\frac{\pi}{4}}e^{\sec^2 x}dx$$
I tried $t=\frac{\pi}{4}-x$ but this seems not work. Any hints?
 A: $\int_0^\frac{\pi}{4}e^{\sec^2 x}~dx$
$=\int_0^\frac{\pi}{4}\sum\limits_{n=0}^\infty\dfrac{\sec^{2n}x}{n!}~dx$
$=\int_0^\frac{\pi}{4}dx+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{\sec^{2n}x}{n!}~dx$
$=[x]_0^\frac{\pi}{4}+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{\sec^{2n-2}x}{n!}~d(\tan x)$
$=\dfrac{\pi}{4}+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{(1+\tan^2x)^{n-1}}{n!}~d(\tan x)$
$=\dfrac{\pi}{4}+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{C_{k-1}^{n-1}\tan^{2k-2}x}{n!}~d(\tan x)$
$=\dfrac{\pi}{4}+\left[\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(n-1)!\tan^{2k-1}x}{n!(n-k)!(k-1)!(2k-1)}\right]_0^\frac{\pi}{4}$
$=\dfrac{\pi}{4}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{1}{n(n-k)!(k-1)!(2k-1)}$
$=\dfrac{\pi}{4}+\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{1}{n(n-k)!(k-1)!(2k-1)}$
$=\dfrac{\pi}{4}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{1}{(n+k+1)n!k!(2k+1)}$
$=\dfrac{\pi}{4}+\int_0^1\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{t^{n+k}}{n!k!(2k+1)}~dt$
$=\dfrac{\pi}{4}+\int_0^1\sum\limits_{k=0}^\infty\dfrac{t^ke^t}{k!(2k+1)}~dt$
$=\dfrac{\pi}{4}+\left[\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(-1)^{k-n}t^ne^t}{n!(2k+1)}\right]_0^1$ (according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Indefinite_integral)
$=\dfrac{\pi}{4}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(-1)^{k-n}e}{n!(2k+1)}-\sum\limits_{k=0}^\infty\dfrac{(-1)^k}{2k+1}$
$=\dfrac{\pi}{4}+\sum\limits_{n=0}^\infty\sum\limits_{k=n}^\infty\dfrac{(-1)^{k-n}e}{n!(2k+1)}-\dfrac{\pi}{4}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^ke}{n!(2n+2k+1)}$
$=\dfrac{e}{2}\Phi_1\left(\dfrac{1}{2},1,\dfrac{3}{2};-1,1\right)$ (according to http://en.wikipedia.org/wiki/Humbert_series)
