TITLE: Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$
QUESTION [16 upvotes]: While I was working on this question by @Vladimir Reshetnikov, I've conjectured the following closed-forms.
$$
I_0(n)=\int_0^\infty \frac{1}{\left(\cosh x\right)^{1/n}} \, dx \stackrel{?}{=} \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\tfrac{1}{2n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{2n}\right)},
$$
for all $n\geq1$ real numbers. In another form:
$$
{_2F_1}\left(\begin{array}c\tfrac{1}{2n},\tfrac1n\\1+\tfrac{1}{2n}\end{array}\middle|\,-1\right) \stackrel{?}{=} \frac{\sqrt{\pi}}{n\,2^{1+\frac1n}} \frac{\Gamma\left(\tfrac{1}{2n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{2n}\right)}.
$$
Another conjectured closed-form is
$$
I_1(n)=\int_0^\infty \frac{1}{\left(1+\cosh x\right)^{1/n}} \, dx \stackrel{?}{=} \frac{\sqrt{\pi}}{2^{1/n}} \frac{\Gamma\left(\tfrac{1}{n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{n}\right)},
$$
for all $n \geq 1$ real numbers. In another form:
$$
{_2F_1}\left(\begin{array}c\tfrac1n,\tfrac2n\\1+\tfrac{1}{n}\end{array}\middle|\,-1\right) \stackrel{?}{=} \frac{\sqrt{\pi}}{n\,2^{\frac2n}} \frac{\Gamma\left(\tfrac{1}{n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{n}\right)}.
$$
Here $\cosh$ is the hyperbolic cosine function, $\Gamma$ is the gamma function, and ${_2F_1}$ is the hypergeometric function.
Questions.

$1^{\text{st}}$ question. How could we prove the conjectured closed-form for $I_0$ and $I_1$.
$2^{\text{nd}}$ question. How could we show the equivalent hypergeometric forms.
$3^{\text{rd}}$ question. There is a closed-form of $I_a(n) = \int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a\geq0,n\geq1$ real numbers in term of Appell $F_1$ function. Could we get a closed-form just in term of gamma function?

REPLY [9 votes]: Although not as simple as nospoon's answer, here's another approach for the first one.
For $s>0$, 
$$ \begin{align} \int_{0}^{\infty} \frac{1}{\cosh^{s}(x)} \, dx &= 2^{s} \int_{0}^{\infty} \frac{1}{(e^{x}+e^{-x})^{s}} \\ &= 2^{s} \int_{0}^{1} \frac{1}{(u^{-1}+u)^{s}} \, \frac{du}{u} \\ &= 2^{s} \int_{0}^{1} \frac{u^{s-1}}{(u^{2}+1)^{s}} \, du \\ &= 2^{s} \int_{0}^{1} \frac{(\sqrt{v})^{s-1}}{(v+1)^{s}} \, \frac{dv}{2 \sqrt{v}}   \\ &=2^{s-1} \int_{0}^{1} \frac{v^{s/2-1}}{(v+1)^{s}} \, dv \\ &=2^{s-1} \, B \left(\frac{s}{2}, 1 \right) \, _2F_1 \left(s,\frac{s}{2}; 1+ \frac{s}{2}; -1 \right) \tag{1}\\ &=2^{s-1} \,  B \left(\frac{s}{2}, 1 \right) \frac{\Gamma \left(1+ \frac{s}{2} \right) \Gamma \left(1+ \frac{s}{2} \right)}{\Gamma \left(1+s \right) \Gamma\left(1 \right)} \tag{2} \\ &= 2^{s-1} \, \Gamma \left(\frac{s}{2}  \right) \frac{\Gamma \left(1+ \frac{s}{2} \right)}{\Gamma(1+s)} \\ &= \frac{\sqrt{\pi}}{2} \frac{ \Gamma (\frac{s}{2})}{\Gamma(\frac{1}{2}+ \frac{s}{2})} \tag{3} \end{align}$$

$(1)$ Euler's integral representation of the Gaussian hypergeometric function
$(2)$ Kummer's theorem
$(3)$ Duplication formula for the gamma function