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


  • $\begingroup$ Where you wrote $2\pi/\alpha\mathrm{csc}\cdots$, the lack of space between $\alpha$ and $\mathrm{csc}$ would not have happened if instead of \mathrm{csc} you had coded it as \csc. I changed it. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 1 '15 at 16:32
  • $\begingroup$ I don't know about the result you mentioned; but you can start by inverting the limits to [0,1/b] then normalize so that it's [0,1] then assign zt=(normalizew)^(-a/2). Then carry through to put it into the form of DLMF 15.6.x. $\endgroup$ – rrogers Mar 8 '16 at 22:06
  • $\begingroup$ Do you want to do contour integration (which probably is intended) or a direct integration using Hypergeometric functions? The last is easy but probably restricting in terms of the signs and values of $\alpha$ $\endgroup$ – rrogers Jul 24 '17 at 18:35

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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