Compute the following integrals.
\(i\) \(\displaystyle \int_0^\infty \frac{1}{(1 + x^n)^2} \, dx\), \(n \geq 1\) (ii) \(\displaystyle \int_0^\infty \frac{\cos x}{(x^2 + a^2)^2} \, dx\), \(a \in \mathbb R\) (iii) \(\displaystyle \int_0^\pi \frac{1}{a + \sin \theta} \, d \theta\), \(a>1\)
\(iv\) \(\displaystyle \int_0^{\frac{\pi}{2}} \frac{d \theta}{a+ \sin ^2 \theta},\) \(a >0\). (v) \(\displaystyle \int_{|z|=2} \frac{1}{(z^{5} -1) (z-3)} \, dz\) (v) \(\displaystyle \int_{- \infty}^{\infty} \frac{\sin \pi a}{\cosh \pi x + \cos \pi a} e^{- i x \xi} \, d x\), \(0< a <1\), \(\xi \in \mathbb R\) (vi) \(\displaystyle \int_{|z| = 1} \cot^2 z \, dz\).
Solution
Reading the seven displayed integrals in order, their values are as follows.
For the first integral, when \(n>1\) the substitution \(u=x^n\) gives \[ \int_0^\infty\frac{dx}{(1+x^n)^2} =\frac1n B\left(\frac1n,2-\frac1n\right) =\boxed{\frac{\pi(n-1)}{n^2}\csc\frac\pi n}. \] For \(n=1\) its value is \(1\).
For the second integral, if \(a=0\) it diverges. For \(a\ne0\), writing \(A=|a|\) and differentiating \[ \int_0^\infty\frac{\cos x}{x^2+A^2}\,dx =\frac{\pi}{2A}e^{-A} \] with respect to \(A\) gives \[ \boxed{ \int_0^\infty\frac{\cos x}{(x^2+a^2)^2}\,dx =\frac{\pi e^{-|a|}(|a|+1)}{4|a|^3}.} \]
For the third integral, \(t=\tan(\theta/2)\) gives \[ 2\int_0^\infty\frac{dt}{at^2+2t+a}, \] so for \(a>1\), \[ \boxed{ \int_0^\pi\frac{d\theta}{a+\sin\theta} =\frac{2}{\sqrt{a^2-1}}\arctan\sqrt{a^2-1}.} \]
For the fourth integral, \(t=\tan\theta\) gives \[ \boxed{ \int_0^{\pi/2}\frac{d\theta}{a+\sin^2\theta} =\frac{\pi}{2\sqrt{a(a+1)}}.} \]
For the fifth integral, all five roots of \(z^5-1\) lie inside \(|z|=2\), while \(z=3\) lies outside. Since the residue at infinity of \[ F(z)=\frac1{(z^5-1)(z-3)} \] is zero and \[ \operatorname{Res}(F;3)=\frac1{242}, \] the sum of residues inside the contour is \(-1/242\). Hence \[ \boxed{ \int_{|z|=2}\frac{dz}{(z^5-1)(z-3)}=-\frac{\pi i}{121}.} \]
For the sixth integral, a rectangular contour of height \(2\pi i\) applied to \[ \frac{e^{-ikz}}{\cosh z+\cos(\pi a)}, \qquad k=\frac\xi\pi, \] gives, for \(\xi\ne0\), \[ \boxed{ \int_{-\infty}^{\infty} \frac{\sin(\pi a)e^{-ix\xi}} {\cosh(\pi x)+\cos(\pi a)}\,dx =\frac{2\sinh(a\xi)}{\sinh\xi}.} \] At \(\xi=0\), the continuous limiting value is \(2a\).
Finally, \[ \cot z=\frac1z-\frac z3+O(z^3), \] so \[ \cot^2z=\frac1{z^2}-\frac23+O(z^2). \] The only pole inside \(|z|=1\) is \(0\), and its residue is zero. Thus \[ \boxed{\int_{|z|=1}\cot^2z\,dz=0.} \]