By using $(71)$ and $(70)$ identities of Mathworld's hypergeometric function page, $${_2}F{_1}(1,1;\tfrac12;\tfrac12)\stackrel{(71)}{=}2\,{_2}F{_1}(1,-\tfrac12;\tfrac12;-1)\stackrel{(70)}{=}2\left(\tfrac{\sqrt\pi}2\tfrac{\Gamma(\tfrac12)}{\Gamma(1)}+1\right)=\pi+2.$$ But, WolframAlpha calculates it as $\tfrac{\pi}2+2.$ Where is my mistake?
I also wonder whether there is an Euler-type integral that can be used to calculate ${_2}F{_1}(1,1;\tfrac12;\tfrac12)$.
Thanks for reading.
On
UPDATE: The identity $${_2}F{_1}(1,1;\tfrac{1}{2};\tfrac{1}{2}) = \frac{\pi}{2}+2$$ can actually be obtained directly from Euler's integral representation. I show this in the second part of my answer.
As Semiclassical pointed out in the comments, equation (70) should be $${_2}F_1(1,-a;a;-1)=\frac{\sqrt{\pi}}{2} \frac{\Gamma(a+1)}{\Gamma(a+\frac{1}{2})}+1. $$
We can use Euler's integral representation to prove that this indentity holds for $a>0$.
First assume that $a>1$.
Using Euler's integral representation, we have $$ \begin{align} {_2}F_1(1,-a;a;-1) &= {_2}F_1(-a,1;a;-1) \\ &= \frac{1}{B(1,a-1)} \int_{0}^{1} x^{1-1} (1-x)^{a-2}(1+x)^a \, \mathrm dx \\&= (a-1) \int_{0}^{\infty} (1-x)^{a-2}(1+x)^{a} \, \mathrm dx. \end{align}$$
Then integrating by parts, we have
$$ \begin{align} {_2}F_1(1,-a;a;-1) &= - (1+x)^{a}(1-x)^{a-1} \bigg|_{0}^{1} + a\int_{0}^{1} (1-x)^{a-1}(1+x)^{a-1} \, \mathrm dx \\ &= 1 +a \int_{0}^{1} (1-x^{2})^{a-1} \, \mathrm dx. \end{align}$$
We now have an integral representation for ${_2}F_1(1,-a;a;-1) $ that converges for $a>0$.
Evaluating the integral, we get $$ \begin{align} {_2}F_1(1,-a;a;-1) &= 1 + \frac{a}{2} \int_{0}^{1} (1-u)^{a-1} u^{-1/2} \, \mathrm du \\ &= 1+ \frac{a}{2} \, B\left(a, \tfrac{1}{2}\right) \\ &= 1 + \frac{a}{2} \frac{\Gamma(a) \sqrt{\pi}}{\Gamma \left(a+ \frac{1}{2} \right)} \\ &= 1 + \frac{\sqrt{\pi}}{2} \frac{\Gamma(a+1) }{\Gamma \left(a+ \frac{1}{2} \right)}. \end{align}$$
We can also obtain the identity $${_2}F{_1}(1,1;\tfrac{1}{2};\tfrac{1}{2}) = \frac{\pi}{2}+2 $$ directly from Euler's integral representation.
First assume that $a>1$.
Then an integral representation for ${_2}F_{1} \left(1,1;a,\tfrac{1}{2}\right)$ is $$ \begin{align} {_2}F_{1} \left(1,1;a,\tfrac{1}{2}\right) &= \frac{1}{B(1,a-1)} \int_{0}^{1} x^{1-1} (1-x)^{a-2} \left(1- \frac{x}{2} \right)^{-1} \, \mathrm dx \\ &= 2 (a-1) \int_{0}^{1}\frac{(1-x)^{a-2}}{2-x} \, \mathrm dx. \end{align}$$
Integrating by parts, we have $$ \begin{align} {_2}F_{1} \left(1,1;a,\tfrac{1}{2}\right) &= -2 \, \frac{(1-x)^{a-1}}{2-x} \bigg|_{0}^{1} + 2 \int_{0}^{1} \frac{(1-x)^{a-1}}{(2-x)^{2}} \, \mathrm dx \\ &=1 + 2 \int_{0}^{1} \frac{(1-x)^{a-1}}{(2-x)^{2}} \, \mathrm dx \end{align}$$
We now have an integral representation for ${_2}F_{1} \left(1,1;a,\tfrac{1}{2}\right) $ that converges for $a>0$.
At $a= \frac{1}{2}$, we have
$$ \begin{align} {_2}F_{1} \left(1,1;\tfrac{1}{2},\tfrac{1}{2}\right) &= 1 + 2 \int_{0}^{1} \frac{(1-x)^{-1/2}}{(2-x)^{2}} \, \mathrm dx \\ &= 1+ 2\int_{0}^{1} \frac{u^{-1/2}}{(1+u)^{2}} \, \mathrm du \\ &= 1 + 2 \int_{0}^{\pi/4} \frac{\left(\tan^{2}(v)\right)^{-1/2}}{\left(1+\tan^{2}(v)\right)^{2}} \, 2 \tan(v) \sec^{2}(v) \, \mathrm dv \\ &=1 + 4 \int_{0}^{\pi/4} \cos^{2}(v) \, \mathrm dv \\ &= 2+ \frac{\pi}{2}. \end{align}$$
The only real Euler-type integral (barring complex contour integrals that look like Euler-type integrals in their integrands) is \begin{align} {_2}F{_1}(1,1;\frac{1}{2}; -1) = 2 \int_0^\infty e^{-2t} {_1}F{_1}(1,\frac{1}{2}, t) \,dt, \end{align}
(NIST Handbook 2010, eq. 13.10.3)
It is easily verifiable that
$${_1}F{_1}(1,\frac{1}{2}, t) = 1 + e^t \sqrt{\pi} \sqrt{t} \;\text{erf}(\sqrt{t})$$
in which $\text{erf}$ is the error function (hint: combine NIST Handbook 2010, eq. 13.6.5 and 13.6.7 p. 332 with $a=-\frac{1}{2}$ and integrate using 13.3.15; or expand series).
Hence:
\begin{align} {_2}F{_1}(1,1;\frac{1}{2}; -1) = 1+ 2 \sqrt{\pi} \; \mathscr{L}(\sqrt{t} \text{ erf}(\sqrt{t}); 1) \end{align}
The Laplace transform $\mathscr{L}(\sqrt{t} \text{ erf}(\sqrt{t}); p)$ is tabulated (for example, Prudnikov et al. Vol. 4, eq. 8 p. 172):
$$\mathscr{L}(\sqrt{t} \text{ erf}(\sqrt{t}); p) = \frac{1}{\sqrt{\pi p^3}}\left(\frac{\sqrt{p}}{p+1} + \arctan(\frac{1}{\sqrt{p}})\right)$$ whence
\begin{align} {_2}F{_1}(1,1;\frac{1}{2}; -1) &= 1+ 2 \sqrt{\pi} \frac{1}{\sqrt{\pi }}\left(\frac{1}{2} + \arctan(1)\right) \\ &= 2 + 2 \arctan(1) \\ &= 2 + \frac{\pi}{2} \end{align}
It follows that Mathworld has a mistake in its equation $(70)$ as commented by Semiclassical.