Rational series representation of $e^\pi$ This question is related to Why $e^{\pi}-\pi \approx 20$, and $e^{2\pi}-24 \approx 2^9$? by Tito Piezas III.
Andrew Fraker (2014) found an almost-integer which is equivalent to the following approximation
$$e^\pi\approx\frac{5^4}{3^3}$$
https://en.wikipedia.org/wiki/Mathematical_coincidence#Containing_both_.CF.80_and_e
However, $e^\pi<\frac{5^4}{3^3}$, so it cannot be obtained by truncating of the following series of rational terms, because the sequence of partial sums is monotonically increasing.
$$
e^\pi=e^{6asin\left(\frac{1}{2}\right)}=\sum_{k=0}^\infty\frac{3\left(e^{3\pi}-\left(-1\right)^ke^{-3\pi}\right)\Gamma\left(\frac{k}{2}+3i\right)\Gamma\left(\frac{k}{2}-3i\right)}{2 \pi k!}
$$
This formula is built from the integer sequence http://oeis.org/A166748 with contributions by Povolotsky, Hasler, Mathar and Kotesovec.
A similar series with slower convergence is obtained from Kotesovec closed form for http://oeis.org/A166741
$$
e^\pi=e^{2asin\left(1\right)}=\sum_{k=0}^\infty\frac{2^{k-1}\left(e^{\pi}-\left(-1\right)^ke^{-\pi}\right)\Gamma\left(\frac{k}{2}+i\right)\Gamma\left(\frac{k}{2}-i\right)}{\pi k!}
$$
Using $\frac{1}{2}$ instead of $1$ as the argument for $asin$, the third root of $e^\pi$ is obtained.
$$
e^\frac{\pi}{3}=e^{2asin\left(\frac{1}{2}\right)}=\sum_{k=0}^\infty\frac{\left(e^{\pi}-\left(-1\right)^ke^{-\pi}\right)\Gamma\left(\frac{k}{2}+i\right)\Gamma\left(\frac{k}{2}-i\right)}{2\pi k!}
$$
For the sixth root of $e^{\pi}$ the argument of the gamma function becomes a single fraction.
$$e^{\frac{\pi}{6}}=e^{asin\left(\frac{1}{2}\right)}=\sum_{k=0}^{\infty}\frac{\left(e^{\frac{\pi}{2}} - (-1)^k e^{-\frac{\pi}{2}}\right)\Gamma\left(\frac{k+i}{2}\right)\Gamma\left(\frac{k-i}{2}\right)}{4\pi k!}$$
Q1: How can Fraker's approximation be derived by truncating a series for a fractional power of $e^\pi$?
Q2: How can integrals related to these series be obtained?
 A: 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.
