Hot Integral - 19

Calculus Level 5

0 cosh ( n x ) cosh ( π x ) d x \large \int_0^\infty \dfrac{\cosh(nx)}{\cosh(\pi x) } \, dx

For constant n < π |n | < \pi , the value of the integral above is equal to

A B sec ( n C D ) \dfrac AB \sec \left ( \dfrac{n^C}D \right)

where A , B , C A,B,C and D D are positive integers with A , B A,B coprime.

Calculate A + B + C + D A+B+C+D .


This is a part of Hot Intergals


The answer is 6.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Ronak Agarwal
Dec 30, 2015

We have I = 0 c o s h ( n x ) c o s h ( π x ) d x \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { cosh(nx) }{ cosh(\pi x) } dx }

Since the integrand is even hence we have :

I = 1 2 c o s h ( n x ) c o s h ( π x ) d x \displaystyle I = \frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ \frac { cosh(nx) }{ cosh(\pi x) } dx }

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

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

I = 1 2 ( 0 y π + n 1 1 + y 2 π d y + 0 y π n 1 1 + y 2 π d y ) \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 = 0 x m 1 d x 1 + x n = π n ( csc m π n ) 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 1 1 + x n = y \dfrac{1}{1+{x}^{n}} = {y} , to get :

J = 1 n 0 1 y m n ( 1 y ) m n 1 d y \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 :

J = 1 n ( Γ ( m n ) Γ ( 1 m n ) Γ ( 1 ) ) = π n csc ( π m n ) \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 :

I = 1 2 ( 1 2 csc ( π + n 2 ) + 1 2 csc ( π n 2 ) ) = 1 2 sec ( n 2 ) \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) }

A nice derivation, but the substitution y = 1 1 + x n y = \tfrac{1}{1+x^n} is not an obvious one!

I would recommend using contour integration for calculating J J . Try the contour shaped like a piece of pie, made up of two radial lines at arguments 0 0 and 2 π n \tfrac{2\pi}{n} , going from radius ϵ \epsilon to radius R R , joined by the arcs of circles at radii ϵ \epsilon and R R .

The integrals of z m 1 1 + z n \tfrac{z^{m-1}}{1+z^n} along the two radial spokes are simply multiples of each other. The integral along the circular arc of radius R R is O ( R m 1 O(R^{m-1} ), and so tends to 0 0 as R R \to \infty . The integral along the circular arc of radius ϵ \epsilon is O ( ϵ m ) O(\epsilon^m) , and so tends to 0 0 as ϵ 0 \epsilon \to 0 . This means that calculating the integral amounts to evaluating the residue at the single pole e π i n e^{\pi i}{n} that lies inside the contour.

It is worth pointing out that J J only exists for 0 < m < 1 0 < m < 1 . Otherwise the two circular arc integrals do not vanish.

Mark Hennings - 5 years, 5 months ago

Log in to reply

I disagree with the fact that y = 1 1 + x n y=\frac{1}{1+x^n} is not an obvious one. Even I did the same substitution. This substitution is not an obvious one for those who don't have any idea about beta function.

Aditya Kumar - 5 years, 5 months ago

Log in to reply

I am well aware of the beta function, having been working with it on and off for the last 30 years or more...

My comment was more to the point that, given the integral for J J , a complex analyst would recognise it as a very standard integral to be evaluated directly by contour integration, whereas the substitution method is more indirect, first converting the integral into a shape which can then be identified using (previously derived) formulae for the gamma and beta functions.

Mark Hennings - 5 years, 5 months ago

Log in to reply

@Mark Hennings I never meant that you were not knowing beta function. I could never think about that also. You are one of the best in calculus in brilliant. I was talking about those who are new to calculus. I don't know complex analysis. So I found your solution more complicated. Sorry to bother you.

Aditya Kumar - 5 years, 5 months ago

Actually I haven't studies complex analysis yet.

Ronak Agarwal - 5 years, 5 months ago
Mark Hennings
Dec 29, 2015

Let's use the right contour! First note that R e n x cosh π x d x = 2 0 cosh n x cosh π x d x \int_{\mathbb{R}} \dfrac{e^{nx}}{\cosh \pi x}\,dx \; = \; 2 \int_0^\infty \dfrac{\cosh nx}{\cosh \pi x}\,dx and integrate e n z cosh π z \dfrac{e^{nz}}{\cosh \pi z} around the rectangular contour γ N \gamma_N with corners ( ± N , 0 ) (\pm N,0) and ( ± N , i ) (\pm N,i) . The only pole of the function inside the contour γ N \gamma_N is a simple one at z = 1 2 i z=\tfrac12i , and so γ N e n z cosh π z d z = 2 π i R e s x = 1 2 i e n z cosh π z = 2 π i e 1 2 n i π sinh 1 2 π i = 2 e 1 2 n i \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 + π i ) = cosh z \cosh(z+\pi i) = -\cosh z for all z z , the integrals along the two long sides of γ N \gamma_N are equal to [ N , N ] e n z cosh π z d z = N N e n x cosh π x d x [ N + i , N + i ] e n z cosh π z d z = e n i N N e n x cosh π x d x \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 0 as N N \to \infty (because n < π |n| < \pi ). Thus we deduce that R e n x cosh π x d x + e n i R e n x cosh π x d x = 2 e 1 2 n i \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 R e n x cosh π x d x = sec 1 2 n , \int_{\mathbb{R}} \dfrac{e^{nx}}{\cosh \pi x}\,dx \; = \; \sec \tfrac12n \;, and hence 0 cosh n x cosh π x d x = 1 2 sec 1 2 n , \int_0^\infty \dfrac{\cosh nx}{\cosh \pi x}\,dx \; = \; \tfrac12 \sec \tfrac12 n \;, giving the answer 1 + 2 + 1 + 2 = 6 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!

very nicely...

Aman Rajput - 5 years, 5 months ago
Aditya Malusare
Dec 28, 2015

This solution uses contour integration.

We focus on the integral γ cosh ( n z ) cosh ( π z ) d z \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 γ cosh ( n z ) cosh ( π z ) d z = 2 π i × all poles Res { cosh ( n z ) cosh ( π z ) } \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, γ cosh ( n z ) cosh ( π z ) d z = ρ ρ cosh ( n x ) cosh ( π x ) d x + arc cosh ( n z ) cosh ( π z ) d z \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 ρ \rho is the radius of the semicircle.

With the given constraint n < π |n| < \pi , the contribution of the arc to the total integral tends to 0 0 , i.e. As ρ \rho \to \infty , arc cosh ( n z ) cosh ( π z ) d z = i 0 π cosh ( n ρ e i ϕ ) cosh ( π ρ e i ϕ ) ρ e i ϕ d ϕ 0 \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 0 cosh ( n x ) cosh ( π x ) d x = 1 2 × 2 π i × all poles Res { cosh ( n z ) cosh ( π z ) } \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 ( π z ) \cosh(\pi z) are on the imaginary axis and are of the form z = i ( 1 2 + m ) z = i\left(\dfrac{1}{2} +m \right) where m Z \in \mathbb{Z} . The residue of each term in the summation is of the form lim z i ( 1 2 + m ) ( z i ( 1 2 + m ) ) cosh ( n z ) cosh ( π z ) = lim z i ( 1 2 + m ) n ( z i ( 1 2 + m ) ) sinh ( n z ) + cosh ( n z ) π sinh ( π z ) = ( 1 ) m cosh ( i n ( m + 1 2 ) ) i π \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, 0 cosh ( n x ) cosh ( π x ) d x = 1 2 × 2 π i × m = 0 ( 1 ) m cosh ( i n ( m + 1 2 ) ) i π \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 ( β ) = e β + e β 2 \cosh(\beta) = \dfrac{e^{\beta} + e^{-\beta}}{2} and summing the above (geometric progression), we get 0 cosh ( n x ) cosh ( π x ) d x = 1 2 sec ( n 2 ) \int_{0}^{\infty} \dfrac{\cosh(nx)}{\cosh(\pi x)} \text{d}x = \dfrac{1}{2}\sec\left(\dfrac{n}{2}\right)

I think you have some problems with your solution. The condition that n < π |n| < \pi ensures that the integrand tends to 0 0 for R e z |\mathrm{Re}z| \to \infty , but has no effect on the behaviour of the integrand near the imaginary axis. If you are going to show that the integral on the semicircular arc tends to zero, there is more to do.

Also, your final series does not converge, I think. Your common ratios are complex numbers of modulus 1 1 .

Mark Hennings - 5 years, 2 months ago

Your solution and Mark Henning's solution are basically the same thing right?

Pi Han Goh - 5 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...