INTEGRAL sinh(ax)cosh(bx)/sinh(cx)dx from 0 to \infty

In this short post we will evaluate the following integral

$\int_0^\infty \frac{\sinh(ax)\cosh(bx)}{\sinh(cx)}\,dx=\frac{\pi}{2c} \left( \frac{\sin\left(\frac{a \pi}{c} \right)}{\cos\left(\frac{a \pi}{c} \right)+\cos\left(\frac{b \pi}{c} \right)}\right)$

$\begin{aligned} I&=\int_0^\infty \frac{\sinh(ax)\cosh(bx)}{\sinh(cx)}\,dx\\ &=\frac12\int_0^\infty\frac{\sinh\left((a+b)x\right)+\sinh\left((a-b)x\right)}{\sinh(cx)}\,dx\\ &=\frac12\int_0^\infty\frac{\sinh\left((a+b)x\right)}{\sinh(cx)}\,dx+\frac12\int_0^\infty\frac{\sinh\left((a-b)x\right)}{\sinh(cx)}\,dx \qquad (*)\\ &=\frac{\pi}{4c} \tan\left(\frac{(a+b) \pi}{2c} \right)+\frac{\pi}{4c} \tan\left(\frac{(a-b) \pi}{2c} \right)\\ &=\frac{\pi}{4c} \left(\frac{\sin\left(\frac{(a+b) \pi}{2c} \right)}{\cos\left(\frac{(a+b) \pi}{2c} \right)}+ \frac{\sin\left(\frac{(a-b) \pi}{2c} \right)}{\cos\left(\frac{(a-b) \pi}{2c} \right)}\right)\\ &=\frac{\pi}{2c} \left(\frac{\sin\left(\frac{(a+b) \pi}{2c} \right)\cos\left(\frac{(a-b) \pi}{2c} \right)+\sin\left(\frac{(a-b) \pi}{2c}\right)\cos\left(\frac{(a+b) \pi}{2c}\right)}{2\cos\left(\frac{(a+b) \pi}{2c} \right)\cos\left(\frac{(a-b) \pi}{2c} \right)}\right) \qquad (**)\\ &=\frac{\pi}{2c} \left( \frac{\sin\left(\frac{a \pi}{c} \right)}{\cos\left(\frac{a \pi}{c} \right)+\cos\left(\frac{b \pi}{c} \right)}\right) \qquad \blacksquare\\ \end{aligned}$

Where in (*) we used the previously established result

$\int_0^\infty \frac{\sinh(ax)}{\sinh(bx)}\,dx=\frac{\pi}{2b} \tan\left(\frac{a \pi}{2b} \right)$

In (**) we made use of the Lemmas in the appendix below.

$\int_0^\infty \frac{\sinh(ax)\cosh(bx)}{\sinh(cx)}\,dx=\frac{\pi}{2c} \left( \frac{\sin\left(\frac{a \pi}{c} \right)}{\cos\left(\frac{a \pi}{c} \right)+\cos\left(\frac{b \pi}{c} \right)}\right)$ (1)

Letting $b \to ib$ and $c=\pi$ in (1) we obtain

$\int_0^\infty \frac{\sinh(ax)\cos(bx)}{\sinh(\pi x)}\,dx=\frac{1}{2} \left( \frac{\sin\left(a \right)}{\cos\left(a \right)+\cosh\left(b \right)}\right)$ (2)

Letting $a \to ia$ and $c=\pi$ in (1) we obtain

$\int_0^\infty \frac{\sin(ax)\cosh(bx)}{\sinh(\pi x)}\,dx=\frac{1}{2} \left( \frac{\sinh\left(a \right)}{\cosh\left(a \right)+\cos\left(b \right)}\right)$ (3)

Lemma 1

$\begin{aligned} \cosh(ax)\sinh(bx)&=\left(\frac{e^{ax}+e^{-ax}}{2}\right)\left(\frac{e^{bx}-e^{-bx}}{2}\right)\\ &=\frac{e^{ax}e^{bx}-e^{ax}e^{-bx}+e^{-ax}e^{bx}-e^{-ax}e^{-bx}}{4}\\ &=\frac{e^{(a+b)x}-e^{(a-b)x}+e^{-(a-b)x}-e^{-(a+b)x}}{4}\\ &=\frac{e^{(a+b)x}-e^{-(a+b)x}}{4}+\frac{e^{-(a-b)x}-e^{(a-b)x}}{4}\\ &=\frac12\sinh\left((a+b)x\right)+\frac12\sinh\left((a-b)x\right) \end{aligned}$

Lemma 2

$\sin(a)+\sin(b)=2 \sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)$

Proof:

We have

$\sin\left(\frac{a}{2}+\frac{b}{2}\right)=\cos\left(\frac a2\right)\sin\left(\frac b2\right)+\cos\left(\frac b2\right)\sin\left(\frac a2\right)$ (A.1)

$\cos\left(\frac{a}{2}-\frac{b}{2}\right)=\cos\left(\frac a2\right)\cos\left(\frac b2\right)+\sin\left(\frac a2\right)\sin\left(\frac b2\right)$ (A.2)

Multiplying (A.1) by (A.2) we obtain

$\begin{aligned} \sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)&=\cos^2\left(\frac a2\right)\cos\left(\frac b2\right)\sin\left(\frac b2\right)+\sin^2\left(\frac b2\right)\cos\left(\frac a2\right)\sin\left(\frac a2\right)+\cos^2\left(\frac b2\right)\cos\left(\frac a2\right)\sin\left(\frac a2\right)+\sin^2\left(\frac a2\right)\cos\left(\frac b2\right)\sin\left(\frac b2\right)\\ &=\cos^2\left(\frac a2\right)\frac{\sin(b)}{2}+\sin^2\left(\frac b2\right)\frac{\sin(a)}{2}+\cos^2\left(\frac b2\right)\frac{\sin(a)}{2}+\sin^2\left(\frac a2\right)\frac{\sin(b)}{2}\\ &=\frac{\sin(a)}{2}\left(\cos^2\left(\frac b2\right)+\sin^2\left(\frac b2\right) \right)+\frac{\sin(b)}{2}\left(\cos^2\left(\frac a2\right)+\sin^2\left(\frac a2\right) \right)\\ &=\frac{\sin(a)+\sin(b)}{2} \qquad \blacksquare \end{aligned}$

Lemma 3

$\sin(a)-\sin(b)=2 \cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$

Proof:

We have

$\cos\left(\frac{a}{2}+\frac{b}{2}\right)=\cos\left(\frac a2\right)\cos\left(\frac b2\right)-\sin\left(\frac a2\right)\sin\left(\frac b2\right)$ (A.3)

$\sin\left(\frac{a}{2}-\frac{b}{2}\right)=\sin\left(\frac a2\right)\cos\left(\frac b2\right)-\cos\left(\frac a2\right)\sin\left(\frac b2\right)$ (A.4)

Multiplying (A.3) by (A.4) we obtain

$\begin{aligned} \cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)&=-\cos^2\left(\frac a2\right)\cos\left(\frac b2\right)\sin\left(\frac b2\right)+\cos^2\left(\frac b2\right)\cos\left(\frac a2\right)\sin\left(\frac a2\right)+\sin^2\left(\frac b2\right)\cos\left(\frac a2\right)\sin\left(\frac a2\right)-\sin^2\left(\frac a2\right)\cos\left(\frac b2\right)\sin\left(\frac b2\right)\\ &=\cos^2\left(\frac b2\right)\frac{\sin(a)}{2}-\cos^2\left(\frac a2\right)\frac{\sin(b)}{2}-\sin^2\left(\frac a2\right)\frac{\sin(b)}{2}+\sin^2\left(\frac b2\right)\frac{\sin(a)}{2}\\ &=\frac{\sin(a)}{2}\left(\cos^2\left(\frac b2\right)+\sin^2\left(\frac b2\right) \right)-\frac{\sin(b)}{2}\left(\cos^2\left(\frac a2\right)+\sin^2\left(\frac a2\right) \right)\\ &=\frac{\sin(a)-\sin(b)}{2} \qquad \blacksquare \end{aligned}$

Lemma 4

$\cos(a)+\cos(b)=2 \cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)$

Proof:

We have

Multiplying (A.2) by (A.3) we obtain

$\begin{aligned} \cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)&=\left(\cos\left(\frac a2\right)\cos\left(\frac b2\right)+\sin\left(\frac a2\right)\sin\left(\frac b2\right) \right)\left(\cos\left(\frac a2\right)\cos\left(\frac b2\right)-\sin\left(\frac a2\right)\sin\left(\frac b2\right) \right)\\ &=\cos^2\left(\frac a2\right)\cos^2\left(\frac b2\right)-\sin^2\left(\frac a2\right)\sin^2\left(\frac b2\right)\\ &=\left(\frac{1+\cos(a)}{2}\right)\left(\frac{1+\cos(b)}{2}\right)-\left(\frac{1-\cos(a)}{2}\right)\left(\frac{1-\cos(b)}{2}\right)\\ &=\frac14\left(1+\cos(b)+\cos(a)+\cos(a)\cos(b)\right)-\frac14\left(1-\cos(b)-\cos(a)+\cos(a)\cos(b)\right)\\ &=\frac{\cos(a)+\cos(b)}{2} \qquad \blacksquare \end{aligned}$

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