<s>[INST] Integral gaussian hypergeometric function How can we define integral with interval $[b,\infty)$
$$
\begin{align}
C(b,\alpha) & = \int_b^\infty \frac{1}{1+w^{\alpha/2}}\,\mathrm{d}w \\[8pt]
& = 2\pi/\alpha \csc(2\pi/\alpha)-b_2 F_1 (1,2/\alpha;(2+\alpha)/\alpha;-b^{\alpha/2})
\end{align}
$$
where $_2F_1(\cdot)$ is the Gaussian hypergeometric function
Can anyone help me step by step with this or give some hint where to start since i'm still studying math
best,
 [/INST] I rename the parameter for convenicence:
$$I(a,b)=\int_b^\infty \frac{dw}{1+w^a}=\frac{1}{a} \int_{b^a}^\infty \frac{u^{1/a-1}du}{1+u}=$$
$$=\frac{1}{a} \int_{0}^\infty \frac{u^{1/a-1}du}{1+u}-\frac{1}{a} \int^{b^a}_0 \frac{u^{1/a-1}du}{1+u}=$$
$$=\frac{1}{a} B\left(\frac{1}{a},1-\frac{1}{a} \right)-\frac{b}{a}\int^{1}_0 \frac{v^{1/a-1}dv}{1+b^a v}=$$
$$=\frac{1}{a} \Gamma \left(\frac{1}{a}\right) \Gamma \left(1-\frac{1}{a}\right)-\frac{b}{a} B\left(\frac{1}{a},1 \right) ~{_2F_1} \left(1,\frac{1}{a};1+\frac{1}{a} ;-b^a\right)=$$

$$=\frac{\pi}{a \sin \frac{\pi}{a}}-b  {_2F_1} \left(1,\frac{1}{a};1+\frac{1}{a} ;-b^a\right)$$

All of this directly follows from the definition of the Beta function and the Euler integral for the Hypergeometric function.
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