Rational series representation of $e^\pi$

Solution 1:

I'll try to obtain a more clear form of the series first, then maybe something about the integral.

Let's work with the series:

$$e^\pi=\sum_{k=0}^\infty\frac{2^{k-1}\left(e^{\pi}-\left(-1\right)^ke^{-\pi}\right)\Gamma\left(\frac{k}{2}+i\right)\Gamma\left(\frac{k}{2}-i\right)}{\pi k!}$$

First we explicitly separate even and odd terms for clarity:

$$e^\pi=\sum_{n=0}^\infty\frac{2^{2n} \sinh (\pi) \Gamma\left(n+i\right)\Gamma\left(n-i\right)}{\pi (2n)!}+ \\ +\sum_{n=0}^\infty\frac{2^{2n+1} \cosh (\pi) \Gamma\left(n+\frac{1}{2}+i\right)\Gamma\left(n+\frac{1}{2}-i\right)}{\pi (2n+1)!}=S_1+S_2$$

Using information from Wikipedia, we can write for $n \geq 1$:

$$\Gamma\left(n+i\right)\Gamma\left(n-i\right)=\frac{\pi}{\sinh \pi} (i)_n (-i)_n=\frac{\pi}{\sinh \pi} \prod_{j=0}^{n-1} \left(j^2+1 \right)$$

$$\Gamma\left(n+\frac{1}{2}+i\right)\Gamma\left(n+\frac{1}{2}-i\right)=\frac{\pi}{\cosh \pi} \left(\frac{1}{2}+i \right)_n \left(\frac{1}{2}-i \right)_n= \\ =\frac{\pi}{\cosh \pi} \prod_{j=0}^{n-1} \left(\left(j+\frac{1}{2} \right)^2+1 \right)$$

Now we obtained explicitly rational series:

$$S_1=1+\sum_{n=1}^\infty\frac{2^{2n} }{(2n)!} \prod_{j=0}^{n-1} \left(j^2+1 \right)$$

$$S_2=2+\sum_{n=1}^\infty\frac{2^{2n+1} }{(2n+1)!} \prod_{j=0}^{n-1} \left(\left(j+\frac{1}{2} \right)^2+1 \right)$$

Explicit form of these series is (easy to obtain by writing the ratio of adjacent terms):

$$S_1={_2 F_1} \left(i,-i; \frac{1}{2};1 \right)$$

$$S_2=2~{_2 F_1} \left(\frac{1}{2}+i,\frac{1}{2}-i; \frac{3}{2};1 \right)$$

Numerical check:

Hypergeometric2F1[I,-I,1/2,1]+2 Hypergeometric2F1[1/2+I,1/2-I,3/2,1]-Exp[Pi]

Wolfram Alpha gives $0$.

We can use integral representations of the Hypergeometric functions to obtain the desired integral form, though it would contain complex exponentials.