Compute \[ \int_0^\infty \frac{x^{50}}{x^{100}+1}\,dx. \]
Solution
1. With\[ t=x^{100}, \]the integral becomes\[ \frac1{100} \int_0^\infty \frac{t^{51/100-1}}{1+t} \,dt. \]
The substitution gives \[ x=t^{1/100} \] and \[ dx = \frac1{100} t^{1/100-1} \,dt. \] Therefore \[ \begin{aligned} \frac{x^{50}}{x^{100}+1}\,dx &= \frac{t^{1/2}}{1+t} \frac1{100} t^{1/100-1} \,dt\\ &= \frac1{100} \frac{t^{51/100-1}}{1+t} \,dt. \end{aligned} \]
2. For\[ 0<a<1, \]set\[ J(a) = \int_0^\infty \frac{t^{a-1}}{1+t} \,dt. \]This improper integral converges.
Near \(0\), the integrand is bounded in absolute value by \[ t^{a-1}, \] whose integral converges because \(a>0\). For \(t\geq1\), \[ \frac{t^{a-1}}{1+t} \leq t^{a-2}, \] whose integral over \([1,\infty)\) converges because \(a-2<-1\).
3. On\[ \CC\sm[0,\infty), \]choose the branch\[ \Log z = \log\abs{z} + i\arg z, \qquad 0<\arg z<2\pi, \]and define\[ z^{a-1} = \exp\bigl((a-1)\Log z\bigr). \]
The slit plane is a domain on which the displayed argument is continuous. Thus \(\Log z\) and hence \(z^{a-1}\) are holomorphic there.
4. Let \(\Gamma_{\varepsilon,R}\) be the positively oriented keyholecontour about the positive real axis, with inner radius \(\varepsilon\) andouter radius \(R\), where\[ 0<\varepsilon<1<R. \]Then\[ \int_{\Gamma_{\varepsilon,R}} \frac{z^{a-1}}{1+z} \,dz = 2\pi i\,e^{i\pi(a-1)}. \]
Inside the keyhole contour, the function \[ \frac{z^{a-1}}{1+z} \] has one pole, at \(z=-1\). On the chosen branch, \[ \arg(-1)=\pi, \] so \[ (-1)^{a-1} = e^{i\pi(a-1)}. \] Hence \[ \operatorname{Res}_{z=-1} \frac{z^{a-1}}{1+z} = e^{i\pi(a-1)}. \] The residue theorem gives the displayed contour integral.
5. The contributions from the inner and outer circular arcs of\(\Gamma_{\varepsilon,R}\) tend to \(0\) as\[ \varepsilon\to0^+ \qquad\text{and}\qquad R\to\infty. \]
On the inner circle \(\abs{z}=\varepsilon\), \[ \abs{ \frac{z^{a-1}}{1+z} } \leq \frac{\varepsilon^{a-1}}{1-\varepsilon}. \] Its length is \(2\pi\varepsilon\), so the absolute value of its contribution is at most \[ \frac{2\pi\varepsilon^a}{1-\varepsilon} \longrightarrow 0. \]
On the outer circle \(\abs{z}=R\), \[ \abs{ \frac{z^{a-1}}{1+z} } \leq \frac{R^{a-1}}{R-1}. \] Its length is \(2\pi R\), so the contribution is at most \[ \frac{2\pi R^a}{R-1}. \] Since \(a<1\), this tends to \(0\) as \(R\to\infty\).
6. In the limit\[ \varepsilon\to0^+, \qquad R\to\infty, \]the two straight portions of the keyhole contour contribute\[ \bigl( 1-e^{2\pi ia} \bigr)J(a). \]
On the upper side of the positive axis, \[ \arg z=0, \] so the contribution tends to \[ J(a). \]
On the lower side, \[ \arg z=2\pi, \] so \[ z^{a-1} = e^{2\pi i(a-1)}t^{a-1} = e^{2\pi ia}t^{a-1}. \] This side is traversed from \(R\) back to \(\varepsilon\), so its limiting contribution is \[ -e^{2\pi ia}J(a). \] Adding the two gives the claim.
7. For \(0<a<1\),\[ \boxed{ J(a) = \frac{\pi}{\sin(\pi a)}. } \]
Letting the two radii tend to their limits in step 4 and using steps <1>5–<1>6 gives \[ \bigl( 1-e^{2\pi ia} \bigr)J(a) = 2\pi i\,e^{i\pi(a-1)}. \] Now \[ 1-e^{2\pi ia} = -2i\,e^{i\pi a}\sin(\pi a), \] while \[ e^{i\pi(a-1)} = -e^{i\pi a}. \] Substitution and cancellation yield \[ J(a) = \frac{\pi}{\sin(\pi a)}. \]
8. The requested integral equals\[ \boxed{ \frac{\pi} {100\cos(\pi/100)}. } \]
Apply step 7 with \[ a=\frac{51}{100}. \] Step 1 gives \[ \begin{aligned} \int_0^\infty \frac{x^{50}}{x^{100}+1} \,dx &= \frac1{100} \frac{\pi} {\sin(51\pi/100)}\\ &= \frac{\pi} {100\cos(\pi/100)}, \end{aligned} \] because \[ \sin\left( \frac\pi2+\frac{\pi}{100} \right) = \cos\left( \frac{\pi}{100} \right). \]
9. Q.E.D.
Step 8 is the requested value.