Hot Integral - 19

We have $\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { cosh(nx) }{ cosh(\pi x) } dx }$

Since the integrand is even hence we have :

$\displaystyle I = \frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ \frac { cosh(nx) }{ cosh(\pi x) } dx }$

Put ${e}^{x}=y$ , to get our integral as :

$\displaystyle I = \frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \frac { { y }^{ n-1 }+{ y }^{ -n-1 } }{ { y }^{ \pi }+{ y }^{ -\pi } } } dy$

$\displaystyle I = \frac { 1 }{ 2 } \left( \int _{ 0 }^{ \infty }{ \frac { { y }^{ \pi +n-1 } }{ 1+{ y }^{ 2\pi } } } dy+\int _{ 0 }^{ \infty }{ \frac { { y }^{ \pi -n-1 } }{ 1+{ y }^{ 2\pi } } } dy \right)$

Lemma

$J = \displaystyle \int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 }dx }{ 1+{ x }^{ n } } } =\frac { \pi }{ n } \left( \csc { \frac { m\pi }{ n } } \right)$

To prove the lemma put $\dfrac{1}{1+{x}^{n}} = {y}$ , to get :

$\displaystyle J = \frac { 1 }{ n } \int _{ 0 }^{ 1 }{ { y }^{ \frac { -m }{ n } }{ (1-y) }^{ \frac { m }{ n } -1 }dy }$

Using the definition of beta function and the gamma relfection formula to get :

$\displaystyle J = \frac { 1 }{ n } \left( \frac { \Gamma \left( \frac { m }{ n } \right) \Gamma \left( 1-\frac { m }{ n } \right) }{ \Gamma (1) } \right) =\frac { \pi }{ n } \csc { \left( \frac { \pi m }{ n } \right) }$

Using the lemma we have :

$\displaystyle I = \frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \csc { \left( \frac { \pi +n }{ 2 } \right) } +\frac { 1 }{ 2 } \csc { \left( \frac { \pi -n }{ 2 } \right) } \right) =\frac { 1 }{ 2 } \sec { \left( \frac { n }{ 2 } \right) }$

Let's use the right contour! First note that $\int_{\mathbb{R}} \dfrac{e^{nx}}{\cosh \pi x}\,dx \; = \; 2 \int_0^\infty \dfrac{\cosh nx}{\cosh \pi x}\,dx$ and integrate $\dfrac{e^{nz}}{\cosh \pi z}$ around the rectangular contour $\gamma_N$ with corners $(\pm N,0)$ and $(\pm N,i)$ . The only pole of the function inside the contour $\gamma_N$ is a simple one at $z=\tfrac12i$ , and so $\int_{\gamma_N} \dfrac{e^{nz}}{\cosh \pi z}\,dz \; = \; 2\pi i\mathrm{Res}_{x=\frac12i} \dfrac{e^{nz}}{\cosh \pi z} \; = \; 2\pi i\dfrac{e^{\frac12ni}}{\pi \sinh \frac12\pi i} \; = \; 2e^{\frac12ni}$ Now since $\cosh(z+\pi i) = -\cosh z$ for all $z$ , the integrals along the two long sides of $\gamma_N$ are equal to $\begin{array}{rcl} \displaystyle\int_{[-N,N]} \dfrac{e^{nz}}{\cosh \pi z}\,dz & = & \displaystyle\int_{-N}^N \dfrac{e^{nx}}{\cosh \pi x}\, dx \\ \displaystyle\int_{[-N+i,N+i]} \dfrac{e^{nz}}{\cosh \pi z}\,dz & = & \displaystyle-e^{ni} \int_{-N}^N \dfrac{e^{nx}}{\cosh \pi x}\,dx \end{array}$ and the integrals along the two short sides of the rectangle tend to $0$ as $N \to \infty$ (because $|n| < \pi$ ). Thus we deduce that $\int_{\mathbb{R}} \dfrac{e^{nx}}{\cosh \pi x}\,dx + e^{ni}\int_{\mathbb{R}} \dfrac{e^{nx}}{\cosh \pi x}\,dx \; =\; 2e^{\frac12 ni}$ so that $\int_{\mathbb{R}} \dfrac{e^{nx}}{\cosh \pi x}\,dx \; = \; \sec \tfrac12n \;,$ and hence $\int_0^\infty \dfrac{\cosh nx}{\cosh \pi x}\,dx \; = \; \tfrac12 \sec \tfrac12 n \;,$ giving the answer $1+2+1+2=6$ .

Quite often, hyperbolic functions will be most conveniently handled by contour integration around rectangular contours such as this one, so that the periodicity properties of the hyperbolic functions can be taken advantage of!

This solution uses contour integration.

We focus on the integral $\displaystyle\int_{\gamma} \dfrac{\cosh(nz)}{\cosh(\pi z)} \text{d}z$ over the semicircle in the upper half of the complex plane whose diameter is the real axis.

Using the method of residues, the following equality holds $\int_{\gamma} \dfrac{\cosh(nz)}{\cosh(\pi z)} \text{d}z = 2 \pi i \times \sum_{\text{all poles}} \text{Res} \left\{\dfrac{\cosh(nz)}{\cosh(\pi z)} \right\}$

Now, $\int_{\gamma} \dfrac{\cosh(nz)}{\cosh(\pi z)} \text{d}z = \int_{-\rho}^{\rho} \dfrac{\cosh(nx)}{\cosh(\pi x)} \text{d}x + \int_{\text{arc}} \dfrac{\cosh(nz)}{\cosh(\pi z)} \text{d}z$ where $\rho$ is the radius of the semicircle.

With the given constraint $|n| < \pi$ , the contribution of the arc to the total integral tends to $0$ , i.e. As $\rho \to \infty$ , $\int_{\text{arc}} \dfrac{\cosh(nz)}{\cosh(\pi z)} \text{d}z = i\int_{0}^{\pi} \dfrac{\cosh(n \rho e^{i \phi})}{\cosh(\pi \rho e^{i \phi})} \rho e^{i \phi}\text{d} \phi \to 0$

Hence, using the statements above and also the fact that the integrand is even, we get $\int_{0}^{\infty} \dfrac{\cosh(nx)}{\cosh(\pi x)} \text{d}x = \frac{1}{2} \times 2 \pi i \times \sum_{\text{all poles}} \text{Res} \left\{\dfrac{\cosh(nz)}{\cosh(\pi z)} \right\}$

Now, we observe the fact that all the zeroes of the function $\cosh(\pi z)$ are on the imaginary axis and are of the form $z = i\left(\dfrac{1}{2} +m \right)$ where m $\in \mathbb{Z}$ . The residue of each term in the summation is of the form $\lim_{z \to i\left(\frac{1}{2} +m \right)} \dfrac{\left(z - i\left(\frac{1}{2} +m \right)\right) \cosh(nz)}{\cosh(\pi z)} = \lim_{z \to i\left(\frac{1}{2} +m \right)} \dfrac{n\left(z - i\left(\frac{1}{2} +m \right)\right)\sinh(nz) + \cosh(nz)}{\pi \sinh(\pi z)} = \dfrac{(-1)^m \cosh\left(in\left(m+\frac{1}{2}\right)\right)}{i \pi}$ The above is calculated using L'Hospital's rule

So, $\int_{0}^{\infty} \dfrac{\cosh(nx)}{\cosh(\pi x)} \text{d}x = \frac{1}{2} \times 2 \pi i \times \sum_{m=0}^{\infty}\dfrac{(-1)^m \cosh\left(in\left(m+\frac{1}{2}\right)\right)}{i \pi}$

Finally, using the definition $\cosh(\beta) = \dfrac{e^{\beta} + e^{-\beta}}{2}$ and summing the above (geometric progression), we get $\int_{0}^{\infty} \dfrac{\cosh(nx)}{\cosh(\pi x)} \text{d}x = \dfrac{1}{2}\sec\left(\dfrac{n}{2}\right)$