Let \(\lambda\in\mathbb R\). Evaluate \[ \int_0^\infty (\operatorname{sech}x)^2\cos(\lambda x)\,dx, \] where \[ \operatorname{sech}x=\frac{2}{e^x+e^{-x}}. \]
Solution
1. For \(\lambda>0\), set\[ F_\lambda(z)\coloneqq \frac{e^{i\lambda z}}{\cosh^2 z}. \]On the rectangle with vertices\[ -R,\quad R,\quad R+i\pi,\quad -R+i\pi, \]the only pole of \(F_\lambda\) is the double pole\[ z_0=\frac{i\pi}{2}. \]
The zeros of \(\cosh z\) are \[ z=\frac{(2k+1)i\pi}{2}, \qquad k\in\ZZ. \] Exactly one of them lies in the strip \[ 0<\operatorname{Im}z<\pi, \] namely \(z_0=i\pi/2\). Since each zero of \(\cosh z\) is simple, \(F_\lambda\) has a double pole there.
2. The residue at \(z_0=i\pi/2\) is\[ \operatorname{Res}_{z=z_0}F_\lambda = -i\lambda e^{-\pi\lambda/2}. \]
Write \[ z=z_0+w. \] Then \[ \cosh(z_0+w)=i\sinh w, \] so \[ F_\lambda(z_0+w) = -e^{-\pi\lambda/2} \frac{e^{i\lambda w}}{\sinh^2w}. \] As \(w\to0\), \[ \frac1{\sinh^2w} = \frac1{w^2}+O(1), \qquad e^{i\lambda w} = 1+i\lambda w+O(w^2). \] Thus the coefficient of \(w^{-1}\) is \[ -i\lambda e^{-\pi\lambda/2}, \] which is the residue.
3. If\[ I(\lambda) \coloneqq \int_{-\infty}^{\infty} \frac{e^{i\lambda x}}{\cosh^2x}\,dx, \]then for \(\lambda>0\),\[ I(\lambda) = \frac{\pi\lambda}{\sinh(\pi\lambda/2)}. \]
Let \(B_R\) denote the integral of \(F_\lambda\) along the bottom edge of the rectangle: \[ B_R = \int_{-R}^{R} \frac{e^{i\lambda x}}{\cosh^2x}\,dx. \] Along the top edge, oriented from \(R+i\pi\) to \(-R+i\pi\), use \[ \cosh(x+i\pi)=-\cosh x \] to obtain \[ \int_{\text{top}}F_\lambda(z)\,dz = -e^{-\pi\lambda}B_R. \]
On either vertical edge, \[ \begin{aligned} \abs{\cosh(\pm R+iy)}^2 &= \cosh^2R\cos^2y+\sinh^2R\sin^2y\\ &= \sinh^2R+\cos^2y\\ &\ge \sinh^2R. \end{aligned} \] Also, for \(0\le y\le\pi\), \[ \abs{e^{i\lambda(\pm R+iy)}}=e^{-\lambda y}\le1. \] Each vertical edge has length \(\pi\), so its integral has absolute value at most \[ \frac{\pi}{\sinh^2R}, \] which tends to zero.
The residue theorem and step 2 therefore give, after letting \(R\to\infty\), \[ \left(1-e^{-\pi\lambda}\right)I(\lambda) = 2\pi i \left(-i\lambda e^{-\pi\lambda/2}\right) = 2\pi\lambda e^{-\pi\lambda/2}. \] Since \[ 1-e^{-\pi\lambda} = 2e^{-\pi\lambda/2}\sinh(\pi\lambda/2), \] the stated formula follows.
4. For \(\lambda>0\),\[ \int_0^\infty (\operatorname{sech}x)^2\cos(\lambda x)\,dx = \frac{\pi\lambda}{2\sinh(\pi\lambda/2)}. \]
The function \(\operatorname{sech}^2x\) is even. Therefore the sine part of \(I(\lambda)\) is odd and integrates to zero, while the cosine part is even. Thus \[ I(\lambda) = 2\int_0^\infty (\operatorname{sech}x)^2\cos(\lambda x)\,dx. \] Apply step 3.
5. The formula extends to every real \(\lambda\), with value \(1\) at\(\lambda=0\):\[ \boxed{ \int_0^\infty (\operatorname{sech}x)^2\cos(\lambda x)\,dx = \begin{cases} \dfrac{\pi\lambda}{2\sinh(\pi\lambda/2)},&\lambda\ne0,\\[3mm] 1,&\lambda=0. \end{cases}} \]
The original integral is an even function of \(\lambda\). The quotient \[ \frac{\pi\lambda}{2\sinh(\pi\lambda/2)} \] is also even, so step 4 gives the formula for \(\lambda<0\) as well. For \(\lambda=0\), \[ \int_0^\infty\operatorname{sech}^2x\,dx = \left[\tanh x\right]_0^\infty =1. \] This also equals the limit of the displayed quotient as \(\lambda\to0\).
6. Q.E.D.
Step 5 gives the requested evaluation for every real \(\lambda\).