Integral $\int_{0}^{\infty}\frac1{\sqrt[\alpha]{1+x^\beta}}dx$ Is there a general answer for the integral of the form:$$\int_{0}^{\infty}\dfrac1{\sqrt[\alpha]{1+x^\beta}}dx$$
 A: All integrals of this form are solved by letting $t=\dfrac1{1+x^b}$ and then recognizing the expression of the beta function in the new integral.
A: Using Mathematica:
Assuming[{1 > \[Alpha] > 0, \[Beta] > 1}, 
 Integrate[1/(1 + x^\[Beta])^(1/\[Alpha]), {x, 0, \[Infinity]}]]

$\dfrac{\Gamma \left(1+\frac{1}{\beta }\right) \Gamma \left(\frac{1}{\alpha
   }-\frac{1}{\beta }\right)}{\Gamma \left(\frac{1}{\alpha }\right)}$
A: Let $$y = f(x;a,b) = (1+x^b)^{-1/a}.$$  Then it is easy to see that $a, b > 0$ is a sufficient (but not necessary) condition for $f(x;a,b)$ to be a strictly decreasing function of $x$; moreover, under this condition, we also have $f(0;a,b) = 1$ and $\displaystyle \lim_{x \to \infty} f(x;a,b) = 0$.  Thus $$I = \int_{x=0}^\infty f(x;a,b) \, dx = \int_{y=0}^1 f^{-1}(y;a,b) \, dy = \int_{y=0}^1 (y^{-a}-1)^{1/b} \, dy = \int_{y=0}^1 y^{-a/b} (1 - y^a)^{1/b} \, dy.$$  Now with the substitution $t = y^a$, $y = t^{1/a}$, $dy = \dfrac{t^{1/a-1}}{a} \, dt$, we obtain $$I = \frac{1}{a} \int_{t=0}^1 t^{1/a-1/b-1} (1-t)^{1/b} \, dt.$$  This is clearly a beta integral:  $$B(m,n) = \int_{t=0}^1 t^{m-1} (1-t)^{n-1} \, dt = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$ for the choice $m = 1/a-1/b$, $n = 1/b+1$, and therefore has value $$I = \frac{\Gamma(\frac{1}{a} - \frac{1}{b})\Gamma(\frac{1}{b}+1)}{a \Gamma(\frac{1}{a}+1)} = \frac{\Gamma(\frac{1}{a} - \frac{1}{b})\Gamma(\frac{1}{b}+1)}{\Gamma(\frac{1}{a})}$$ where the last equality is due to the well-known property $\Gamma(z+1) = z\Gamma(z)$.  The restrictions on $a, b$ for convergence of the integral is $0 < a < b$.
