UC Berkeley Integral Problem: Show that $\int_0^{2\pi} \frac{\min(\sin x, \cos x)}{\max(e^{\sin x},e^{\cos x})}\ {\rm d}x = -4\sinh(1/{\sqrt2})$.

Question

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!

Answer 1

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.

Answer 2

If we call $f_1(x)=\min(\sin x,\cos x),\,f_2(x)=\max(e^{\sin x},e^{\cos x})$ then we see that: $$f_1(x)=\sin(x) \{0\le x\le \frac{\pi}4,\frac{5\pi}4\le x\le2\pi\}$$ $$f_1(x)=\cos(x)\{\frac{\pi}4\le x\le\frac{5\pi}4\}$$ $$f_2(x)=e^{\cos x}\{0\le x\le\frac{\pi}4,\frac{5\pi}4\le x\le2\pi\}$$ $$f_2(x)=e^{\sin x}\{\frac{\pi}4\le x\le \frac{5\pi}4\}$$ and so: $$\int_0^{2\pi}\frac{\min(\sin x,\cos x)}{\max(e^{\sin x},e^{\cos x})}dx=\int_0^{\pi/4}\frac{\sin(x)}{e^{\cos x}}dx+\int_{\pi/4}^{5\pi/4}\frac{\cos(x)}{e^{\sin x}}dx+\int_{5\pi/4}^{2\pi}\frac{\sin(x)}{e^{\cos x}}dx$$ And this is now easy to solve using simple $u$ substitution

Answer 3

WE will break the integral $I$ in four parts parts: $$I_1=\int_{0}^{\pi/4} \sin x ~e^{-\cos x}~ dx=-\int_{1}^{1/\sqrt{2}} e^{-t} dt=e^{-1/\sqrt{2}}-e^{-1}.$$ $$I_2=\int_{\pi/4}^{\pi/2} \cos x~e^{-\sin x}~ dx=\int_{1/\sqrt{2}}^{1} e^{-t} dt=-e^{-1}+e^{-1/\sqrt{2}}$$ $$I_3=\int_{\pi/2}^{5\pi/4} \cos x ~ e^{-\sin x} ~dx=e^{-1}-e^{1/\sqrt{2}}$$ $$I_4=\int_{5\pi/4}^{2\pi} \sin x ~ e^{-\cos x}~dx =e^{-1}-e^{1/\sqrt{2}} $$ Addinfg all four we get $$I=I_1+I_2+I_3+I_4=2(e^{-1/\sqrt{2}}-e^{1sqrt{2}}]=-4 \sinh(1/\sqrt{2}).$$