Integral $\int_0^\infty \frac{\sin a x-\sin b x}{\cosh \beta x}\frac{dx}{x}$

Question

Hi I am trying to prove this interesting integral $$ \mathcal{I}:=\int_0^\infty \frac{\sin a x-\sin b x}{\cosh \beta x}\frac{dx}{x}=2\arctan\left(\frac{\exp(\frac{a\pi}{2\beta})-\exp(\frac{b\pi}{2\beta})}{1+\exp{\big(\frac{(a+b)\pi}{2\beta}}\big)}\right), \qquad Re(\beta)>0. $$ This is related to another integral posted except that had cosines in the numerator instead of sine functions. The result is quite different, very interesting integrals. It sure looks like a Frullani but this one's result is quite different looking than just a logarithm function..I

A Frullani integral is $$ \int_0^\infty \frac{f(ax)-f(bx)}{x}dx=\big[f(0)-f(\infty)\big]\log \frac{b}{a}. $$ In this example, $\mathcal{I}$ has a term $\cosh \beta x$ in the denominator which it makes it appear different. Partial integration and splitting the integral up didn't work and ran into convergent issues. How can we solve this integral? Thank you.

Note $$ 2\cosh x=e^x+e^{-x}. $$

Answer 1

Consider the integral \begin{align} \int_{b}^{a} \cos(\mu x) \ d\mu = \left[ \frac{\sin(x \mu)}{x} \right]_{b}^{a} = \frac{\sin(ax) - \sin(bx)}{x}. \end{align} This integral wil be used in the calculation of the integral in question. The integral to calculate is \begin{align} I = \int_{0}^{\infty} \frac{\sin(ax) - \sin(bx)}{x \ \cosh(\beta x)} \ dx \end{align} and can be seen to be \begin{align} I &= \int_{0}^{\infty} \frac{\sin(ax) - \sin(bx)}{x \ \cosh(\beta x)} \ dx \\ &= \int_{b}^{a} \left( \int_{0}^{\infty} \frac{\cos(\mu x)}{\cosh(\beta x)} \ dx \right) \ d\mu. \end{align} The inner integral can be quickly calculated by using the known integral \begin{align} \int_{0}^{\infty} \frac{\cosh(\alpha x)}{\cosh(\beta x)} \ dx = \frac{\pi}{2 \beta} \ \sec\left( \frac{\alpha \pi}{2 \beta} \right). \end{align} By letting $\alpha = i \mu$ this becomes \begin{align} \int_{0}^{\infty} \frac{\cos(\mu x)}{\cosh(\beta x)} \ dx = \frac{\pi}{2 \beta} \ sech\left( \frac{\mu \pi}{2 \beta} \right) \end{align} and leads to \begin{align} I &= \frac{\pi}{2 \beta} \ \int_{b}^{a} sech\left( \frac{\pi \mu}{2 \beta} \right) \ d\mu \\ &= \int_{(b\pi/2\beta)}^{(a\pi/2\beta)} sech(x) \ dx \\ &= 2 \left[ \tan^{-1}\left(e^{\frac{a\pi}{2\beta}} \right) - \tan^{-1}\left(e^{\frac{b\pi}{2\beta}} \right) \right] \end{align} which can be restated as \begin{align} \int_{0}^{\infty} \frac{\sin(ax) - \sin(bx)}{x \ \cosh(\beta x)} \ dx = 2 \tan^{-1}\left( \frac{e^{a\pi/2\beta} - e^{b\pi/2\beta}}{1 + e^{(a+b)\pi/2\beta}}\right). \end{align}

Answer 2

First do a substitution $A = a/\beta$, $B = b/\beta$ to reduce the integral to $$ \int_0^{\infty} \frac{\sin A x}{x \cosh x} dx -\int_0^{\infty} \frac{\sin B x}{x \cosh x} dx $$ These integrals were calculated in a previous question, and the result is

$$ \left(2 \arctan e^{A\pi/2} -\frac{\pi}{2}\right)- \left(2\arctan e^{B\pi/2} -\frac{\pi}{2}\right)= 2 \left( \arctan e^{A\pi/2} - \arctan e^{B\pi/2} \right) $$

By the tangent addition formula, we have $\tan (a-b)=\frac{\tan a - \tan b}{1+\tan a \tan b} $, hence $$\arctan a -\arctan b = \arctan \frac{a-b}{1+ab} $$ and this gives the desired result. (By the way, I wonder why someone would intentionally write the answer in such a complicated form!)

Answer 3

Feynman’s trick

Let $$ I(a)=\int_0^{\infty} \frac{\sin (a x)-\sin (b x)}{x \cosh (\beta x)} d x $$ Differentiating w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{\cos (\alpha x)}{\cosh (\beta x)} d x \\ & =\frac{\pi}{2 \beta} \operatorname{sech}\left(\frac{a \pi}{2 \beta}\right) \end{aligned} $$ where the last answer comes from the post. $$ \begin{aligned} I(a)&=I(a)-I(b)\\ & =\frac{\pi}{2 \beta} \int_b^a \operatorname{sech}\left(\frac{x\pi}{2 \beta}\right) d x \\ & =\int _\frac{b \pi}{2 \beta} ^\frac{a\pi}{2 \beta} \operatorname{sech} x d x \\ & =\int_{\frac{b \pi}{2 \beta}}^{\frac{a \pi}{2 \beta}} \frac{1}{e^{x}+e^{-x}} d x \\ & =\int_{\frac{b \pi}{2 \beta}}^{\frac{a \pi}{2 \beta}} \frac{d\left(e^x\right)}{e^{2 x}+1} \\ & =\tan ^{-1}\left(e^{\frac{a \pi}{2 \beta}}\right)-\tan ^{-1}\left(e^{\frac{b \pi}{2 \beta}}\right) \end{aligned} $$