Trigonometrica Hyperbolica

Calculus Level 5

X = ∫ 0 ∞ cos ⁡ ( π x 2016 ) cosh ⁡ ( π x ) sin ⁡ ( π x 2 ) d x Y = ∫ 0 ∞ cos ⁡ ( π x 2016 ) cosh ⁡ ( π x ) cos ⁡ ( π x 2 ) d x \large{ \begin{aligned} \text{X} & = \int_{0}^{\infty} \dfrac{\cos \left(\frac{\pi x}{2016} \right)}{\cosh (\pi x)} \sin (\pi x^2) \, \mathrm{d}x \\ \\ \text{Y} & = \int_{0}^{\infty} \dfrac{\cos \left(\frac{\pi x}{2016} \right)}{\cosh (\pi x)} \cos (\pi x^2) \, \mathrm{d}x \end{aligned}}

The value of X Y \dfrac {\text{X}}{\text{Y}} can be expressed as tan ⁡ [ π M ( 1 − 1 N 2 ) ] \tan \left[ \dfrac \pi {\text{M}} \left( 1 - \dfrac1{\text{N}^2} \right) \right] , where M \text{M} and N \text{N} are positive integers. Find M × N \text{M}\times \text{N} .

Notation: cosh ⁡ ( x ) = e x + e − x 2 \cosh(x) = \dfrac{e^x + e^{-x}}2 denotes hyperbolic cosine function .


The answer is 16128.

Ishan Singh by Ishan Singh , 22, India

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.

1 solution

Mark Hennings
Oct 28, 2016

Dodging some messy contour integration, the integrals X X and Y Y can be found in Gradshteyn & Ryzkik: X ( c ) = ∫ 0 ∞ cos ⁡ c π x sin ⁡ π x 2 cosh ⁡ π x d x = cos ⁡ 1 4 c 2 π − 1 2 2 cosh ⁡ 1 2 c π Y ( c ) = ∫ 0 ∞ cos ⁡ c π x cos ⁡ π x 2 cosh ⁡ π x d x = sin ⁡ 1 4 c 2 π + 1 2 2 cosh ⁡ 1 2 c π \begin{array}{rcl} \displaystyle X(c) & = & \displaystyle \int_0^\infty \frac{\cos c \pi x\, \sin \pi x^2}{\cosh \pi x}\,dx \; =\; \frac{\cos\frac14c^2\pi - \frac{1}{\sqrt{2}}}{2\cosh \frac12 c \pi} \\ \displaystyle Y(c) & = & \displaystyle \int_0^\infty \frac{\cos c \pi x\, \cos \pi x^2}{\cosh \pi x}\,dx \; = \; \frac{\sin \frac14c^2\pi + \frac{1}{\sqrt{2}}}{2\cosh\frac12 c \pi} \end{array} and so X ( c ) Y ( c ) = cos ⁡ 1 4 c 2 π − cos ⁡ 1 4 π sin ⁡ 1 4 c 2 π + sin ⁡ 1 4 π = − 2 sin ⁡ 1 8 ( c 2 + 1 ) π sin ⁡ 1 8 ( c 2 − 1 ) π 2 sin ⁡ 1 8 ( c 2 + 1 ) π cos ⁡ 1 8 ( c 2 − 1 ) π = tan ⁡ 1 8 ( 1 − c 2 ) π \frac{X(c)}{Y(c)} \;= \; \frac{ \cos \frac14 c^2 \pi - \cos \frac14\pi}{\sin \frac14c^2\pi + \sin\frac14\pi} \; = \; \frac{-2 \sin \frac18(c^2+1)\pi\,\sin\frac18(c^2-1)\pi}{2\sin\frac18(c^2+1)\pi\,\cos\frac18(c^2-1)\pi} \; = \; \tan \tfrac18(1-c^2)\pi and hence a b = 1 8 ( 1 − 1 201 6 2 ) \frac{a}{b} \; = \; \frac18\left(1 - \frac{1}{2016^2}\right) making the answer 8 × 2016 = 16128 8 \times 2016 \; = \; \boxed{16128}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution, you even evaluated both integrals! I did it by first proving that ∫ 0 ∞ cos ⁡ 2 n x cosh ⁡ π x d x = 1 2 cosh ⁡ n \displaystyle \int_0^\infty \dfrac{\cos 2 nx}{\cosh \pi x} \ \mathrm{d}x = \dfrac{1}{2 \cosh n} and substituting this expression for 1/(cosh) in the integral and interchanging the integrals. In this way we can prove that X = tan ⁡ ( π 8 ( 1 − c 2 ) ) Y \text{X} = \tan \left( \dfrac{\pi}{8} (1-c^2) \right) \text{Y} . I'm still searching for an approach to solve for X \text{X} and Y \text{Y} explicitly using Real Analysis though.

Ishan Singh - 4 years, 7 months ago

Log in to reply

OH wow, very nice!

Calvin Lin Staff - 4 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...