contest-mathdefinite integralsintegrationmaxima-minimareal-analysis
Show that $$\int_0^{2\pi} \frac{\mathrm{min}(\sin{x},\, \cos{x})}{\mathrm{max}\left(e^{\sin{x}},\, e^{\cos{x}}\right)}\ \mathrm{d}x = -4\sinh\left(\frac{1}{\sqrt{2}}\right).$$
this problem comes from the 2020 UC Berkeley Integration Bee and was not solved by either of the contestants. Any hints? My initial approach was to compute the maximum and minimum of the specified function by observing the graph for $x\in (0, 2\pi)$ but could not get very far.
Thank you!
Integration by parts (just to simplify the evaluation) gives $$\int_0^\infty\frac{\sinh ax\sinh bx}{(\cosh ax+\cos t)^2}\,dx=\frac{b}{a}\int_0^\infty\frac{\cosh bx\,dx}{\cosh ax+\cos t}.$$ This can be evaluated using the residue theorem. Consider $I_R=\displaystyle\int_{C_R}\frac{e^{bz}\,dz}{\cosh az-\cos t}$ where $C_R$ is the rectangular contour with vertices at $\pm R\pm i\pi/a$; the integrand has simple poles at $z=\pm it/a$, thus $$I_R=2\pi i\left(\operatorname*{Res}_{z=it/a}+\operatorname*{Res}_{z=-it/a}\right)\frac{e^{bz}}{\cosh az-\cos t}=\frac{4\pi i}{a}\frac{\sin(bt/a)}{\sin t},$$ and on the other hand, $\lim\limits_{R\to\infty}I_R$ is equal to $$\int_{-\infty}^\infty\frac{e^{b(x-i\pi/a)}\,dx}{-\cosh ax-\cos t}+\int_\infty^{-\infty}\frac{e^{b(x+i\pi/a)}\,dx}{-\cosh ax-\cos t}=4i\sin\frac{b\pi}{a}\int_0^\infty\frac{\cosh bx\,dx}{\cosh ax+\cos t}.$$
I will concentrate on this part of your proof, the calculation of this integral (the other steps of your proof are correct):
$$\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx$$
We will make use of the Lobachevsky integral formula:
Let $f(x)$ a $\pi$-periodic function (continous or integrable over its period). with
$\displaystyle f(x+\pi) = f(x)$ and $f(\pi-x) = f(x)$, for $0\leq x <\infty$. Then
$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} f(x) dx = \int_{0}^{\frac{\pi}{2}} f(x) dx. $
Now, for your integral: If you make the change of variable $\displaystyle x=\frac{w}{2}$
$\displaystyle I=\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \int_{0}^{\infty} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{w} dw $
Multiplying and dividing by $\sin w$:
$$I= \int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw$$
The function
$$f(w) = \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}$$
is $\pi$-periodic
By the Lobachevsky integral formula:
$$\displaystyle I=\int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw $$
Note that for $w\in\left(0,\frac{\pi}{2}\right)$
$$\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big|= \sin\left(\frac{w}{2}+\frac{\pi}{4}\right)$$
$$\Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|= -\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)$$
Hence
$$\displaystyle\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big| = \sqrt{2}\sin\left(\frac{w}{2}\right)\displaystyle $$ Therefore
$$ I = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw$$
Using $\displaystyle \frac{\sin\left(\frac{w}{2}\right)}{\sin w} = \frac{1}{2}\sec\left(\frac{w}{2}\right)$
we have
$$I = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw = \frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}} \sec\left(\frac{w}{2} \right)dw = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)$$
We can conclude
$$\boxed{\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)}$$
Therefore
$$\boxed{\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx = \frac{\pi}{2} +\operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)} $$
Some numerical approximations from Wolfram:
The sum is 2.44206
while
$$\frac{\pi}{2} + \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)\approx 2.45216..$$
Best Answer
Because the function is periodic, the integral over any interval of length $2 \pi$ leads to the same result. With that said, rewrite this as $$ \int_{\pi/4}^{9 \pi /4} f(x)\,dx = \int_{\pi/4}^{5\pi/4} f(x)\,dx + \int_{5 \pi/4}^{9\pi/4} f(x)\,dx\\ = \int_{\pi/4}^{5 \pi/4} \frac{\cos(x)}{e^{\sin(x)}}\,dx + \int_{\pi/4}^{5 \pi/4} \frac{\sin(x)}{e^{\cos(x)}}\,dx\\ = \int_{\pi/4}^{5 \pi/4} e^{- \sin(x)}\cos(x)\,dx + \int_{\pi/4}^{5 \pi/4} e^{- \cos(x)}\sin(x)\,dx. $$ The integrals can be handled separately, via $u$-substitution.