Let \(0<a<4\) and evaluate \[ \int_0^\infty \frac{x^{\alpha-1}}{1+x^3} ~dx \]
The hypothesis names \(a\) while the integrand has exponent \(\alpha-1\). With \(\alpha=a\), the integral converges only for \(0<a<3\). The stated range \(0<a<4\) is the range of convergence of \(\int_0^\infty x^{a-1}/(1+x^4)\,dx\), and the substitution \(t=x^4\) gives \[\int_0^\infty\frac{x^{a-1}}{1+x^4}\,dx=\frac{\pi}{4}\csc\frac{\pi a}{4}\qquad(0<a<4).\]
Solution
Read the exponent as \(\alpha=a\).
1. The integral \(\int_0^\infty x^{a-1}/(1+x^3)\,dx\) converges if and only if \(0<a<3\).
Near \(0\) the integrand is asymptotic to \(x^{a-1}\), which is integrable on \((0,1)\) if and only if \(a>0\). Near \(\infty\) it is asymptotic to \(x^{a-4}\), which is integrable on \((1,\infty)\) if and only if \(a<3\).
2. For \(0<a<3\),\[\int_0^\infty {x^{a-1}\over1+x^3}\,dx = \boxed{{\pi\over3}\csc{\pi a\over3}}.\]
With \(t=x^3\), \[\int_0^\infty {x^{a-1}\over1+x^3}\,dx ={1\over3}\int_0^\infty {t^{a/3-1}\over1+t}\,dt ={1\over3}B\!\left(\frac a3,1-\frac a3\right) ={1\over3}\Gamma\!\left(\frac a3\right)\Gamma\!\left(1-\frac a3\right),\] and Euler’s reflection formula \(\Gamma(s)\Gamma(1-s)=\pi/\sin(\pi s)\) gives the value.
3. Q.E.D.
Step 2 evaluates the integral on its range of convergence from step 1; for \(3\le a<4\) it diverges.