Compute the following integrals.
\(\displaystyle \int_0^\infty \frac{1}{(1 + x^n)^2} \, dx\), \(n \geq 1\)
\(\displaystyle \int_0^\infty \frac{\cos x}{(x^2 + a^2)^2} \, dx\), \(a \in \mathbb R\)
\(\displaystyle \int_0^\pi \frac{1}{a + \sin \theta} \, d \theta\), \(a>1\)
\(\displaystyle \int_0^{\frac{\pi}{2}} \frac{d \theta}{a+ \sin ^2 \theta},\) \(a >0\).
\(\displaystyle \int_{|z|=2} \frac{1}{(z^{5} -1) (z-3)} \, dz\)
\(\displaystyle \int_{- \infty}^{\infty} \frac{\sin \pi a}{\cosh \pi x + \cos \pi a} e^{- i x \xi} \, d x\), \(0< a <1\), \(\xi \in \mathbb R\)
\(\displaystyle \int_{|z| = 1} \cot^2 z \, dz\).
Solution
We evaluate the seven integrals in order.
For (i), first assume \(n>1\) and put \(u=x^n\). Then \[ \int_0^\infty\frac{dx}{(1+x^n)^2} =\frac1n\int_0^\infty\frac{u^{1/n-1}}{(1+u)^2}\,du =\frac1n B\left(\frac1n,2-\frac1n\right). \] Using \(B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q)\), the recurrence for \(\Gamma\), and Euler’s reflection formula, \[ \boxed{ \int_0^\infty\frac{dx}{(1+x^n)^2} =\frac{\pi(n-1)}{n^2}\csc\frac\pi n, \qquad n>1.} \] For \(n=1\) the integral is directly \[ \int_0^\infty\frac{dx}{(1+x)^2}=1. \]
For (ii), if \(a=0\) the integral diverges at \(0\). If \(a\ne0\), put \(A=|a|\). The standard residue computation gives \[ I(A):=\int_0^\infty\frac{\cos x}{x^2+A^2}\,dx =\frac{\pi}{2A}e^{-A}. \] Differentiating with respect to \(A\), \[ I'(A)=-2A\int_0^\infty\frac{\cos x}{(x^2+A^2)^2}\,dx. \] Therefore \[ \boxed{ \int_0^\infty\frac{\cos x}{(x^2+a^2)^2}\,dx =\frac{\pi e^{-|a|}(|a|+1)}{4|a|^3}, \qquad a\ne0.} \]
For (iii), use \(t=\tan(\theta/2)\). Then \(\sin\theta=2t/(1+t^2)\) and \(d\theta=2\,dt/(1+t^2)\), so \[ \int_0^\pi\frac{d\theta}{a+\sin\theta} =2\int_0^\infty\frac{dt}{at^2+2t+a}. \] Writing \(\Delta=a^2-1>0\) and completing the square gives \[ \boxed{ \int_0^\pi\frac{d\theta}{a+\sin\theta} =\frac{2}{\sqrt{a^2-1}} \arctan\sqrt{a^2-1}, \qquad a>1.} \]
For (iv), set \(t=\tan\theta\). Since \(\sin^2\theta=t^2/(1+t^2)\) and \(d\theta=dt/(1+t^2)\), \[ \int_0^{\pi/2}\frac{d\theta}{a+\sin^2\theta} =\int_0^\infty\frac{dt}{a+(a+1)t^2} =\boxed{\frac{\pi}{2\sqrt{a(a+1)}}}. \]
For (v), the contour \(|z|=2\) encloses all five roots of \(z^5-1\) and excludes \(z=3\). Since \[ F(z)=\frac1{(z^5-1)(z-3)}=O(z^{-6}), \] the residue at infinity is \(0\), so the sum of all finite residues is \(0\). The residue at \(z=3\) is \[ \operatorname{Res}(F;3)=\frac1{3^5-1}=\frac1{242}. \] Thus the sum of the residues inside \(|z|=2\) is \(-1/242\), and \[ \boxed{ \int_{|z|=2}\frac{dz}{(z^5-1)(z-3)} =-\frac{\pi i}{121}.} \]
For (vi), put \(A=\pi a\) and scale \(y=\pi x\). It is enough to evaluate \[ J(k)=\int_{-\infty}^{\infty} \frac{e^{-iky}}{\cosh y+\cos A}\,dy, \qquad k=\frac\xi\pi. \] Assume first \(k>0\) and integrate \[ F(z)=\frac{e^{-ikz}}{\cosh z+\cos A} \] over the rectangle with vertices \(-R,R,R-2\pi i,-R-2\pi i\), oriented clockwise. The vertical sides vanish as \(R\to\infty\). Because the denominator is \(2\pi i\)-periodic, the lower horizontal side contributes \(-e^{-2\pi k}J(k)\). The poles in the strip \(-2\pi<\Im z<0\) are \[ z_1=-i(\pi-A), \qquad z_2=-i(\pi+A). \] Their residue sum is \[ \frac{2i e^{-\pi k}\sinh(Ak)}{\sin A}. \] The clockwise residue theorem therefore gives \[ (1-e^{-2\pi k})J(k) =\frac{4\pi e^{-\pi k}\sinh(Ak)}{\sin A}, \] and hence \[ J(k)=\frac{2\pi}{\sin A}\frac{\sinh(Ak)}{\sinh(\pi k)}. \] The same formula holds for \(k<0\) by evenness, and its limit at \(k=0\) is \(2A/\sin A\). Restoring the scale and the factor \(\sin(\pi a)\) gives \[ \boxed{ \int_{-\infty}^{\infty} \frac{\sin(\pi a)e^{-ix\xi}}{\cosh(\pi x)+\cos(\pi a)}\,dx =\frac{2\sinh(a\xi)}{\sinh\xi}} \] for \(\xi\ne0\), with limiting value \(2a\) at \(\xi=0\).
For (vii), the only pole of \(\cot^2z\) inside \(|z|=1\) is at \(0\). Since \[ \cot z=\frac1z-\frac z3+O(z^3), \] we have \[ \cot^2z=\frac1{z^2}-\frac23+O(z^2), \] whose residue at \(0\) is \(0\). Therefore \[ \boxed{\int_{|z|=1}\cot^2z\,dz=0.} \]