Closed form expression for the integral

Question

Can someone help me find the integral of the following

$$f = \exp\Big(-\pi\lambda T^{2/\beta}r^2\int^\infty_0\frac{1}{1+v^{\beta/2}}dv\Big)$$

It is from the papers and the authors have posted the results in form of some $\Gamma$ function

The final result by authors is

$$ = \exp(-\lambda r^2T^{2/\beta }K(\beta))$$ and $K(\beta) = \frac{2\pi \Gamma(2/\beta)\Gamma(1-2/\beta)}{\beta}$

Serious EDIT: Actually as someone below mentioned there was a typo in the power of $v$ in the denominator. Not it is fixed. Extremely sorry for the trouble to those who tried the solution.

Answer 1

Using the Beta function integral: $$ \begin{align} \int_0^\infty\frac{\mathrm{d}v}{1+v^{\beta/2}} &=\frac2\beta\int_0^\infty\frac{v^{\frac2\beta-1}\,\mathrm{d}v}{1+v}\\ &=\frac2\beta\frac{\Gamma\left(\frac2\beta\right)\Gamma\left(1-\frac2\beta\right)}{\Gamma(1)}\\[6pt] &=\frac2\beta\Gamma\left(\frac2\beta\right)\Gamma\left(1-\frac2\beta\right) \end{align} $$ Note that we are using the formula from that Wikipedia page that says

$$ \operatorname{B}(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\,\mathrm{d}t $$

This would mean that $$ K(\beta)=\frac{2\pi}\beta\Gamma\left(\frac2\beta\right)\Gamma\left(1-\frac2\beta\right) $$

Answer 2

$$\int^\infty_0\frac{1}{1+v^{2/\beta}}\,dv=\frac{\pi \beta}{2} \csc \left(\frac{\pi \beta }{2}\right)\qquad \text{if} \qquad 0<\Re(\beta )<2$$ By Euler's reflection formula $$ \Gamma \left(1-z\right) \Gamma \left(z\right)=\frac \pi {\sin(\pi z)}=\pi \csc (\pi z)$$ Use $z=\frac 2 \beta$ to get $$K(\beta)=\frac{2 \pi}\beta \Gamma \left(1-\frac{2}{\beta }\right) \Gamma \left(\frac{2}{\beta }\right)=\frac{\pi \beta}{2} \csc \left(\frac{\pi \beta }{2}\right)\qquad \text{if} \qquad 0<\Re(\beta )<2$$ I seems that a $\pi$ is missing somewhere in $f$.