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Given this integral

$$\int_{-\infty}^{+\infty}\sin(\cosh x)\cos(\sinh x)\mathrm dx={\pi\over 2}\tag1$$

My try:

Recall $$2\sin(A)\cos(B)=\sin(A-B)+\sin(A+B)$$

$(1)$ becomes

$$\frac12\int_{-\infty}^{+\infty}\cos(\cosh x-\sinh x)+\sin(e^x)\mathrm dx\tag2$$

Recall $$\cosh^2 x-\sinh^2 x=1$$

$(2)$ becomes

$$\frac12\int_{-\infty}^{+\infty}\sin(e^{-x})+\sin(e^x)\mathrm dx\tag3$$

How may be prove $(1)?$

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    $\begingroup$ You should have a $\sin{(A-B)}$ on the RHS of the equation above (2). $\endgroup$ – Chappers May 5 '17 at 17:08
  • $\begingroup$ using @Chappers correction and substitutions $e^{\pm x}=y$ this should be simple enough to finish $\endgroup$ – tired May 5 '17 at 17:12
  • $\begingroup$ you also missed a factor of $2$...# $\endgroup$ – tired May 5 '17 at 17:13
  • $\begingroup$ btw. what went wrong the last week?! $\endgroup$ – tired May 5 '17 at 17:14
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    $\begingroup$ I promised to buy back the reps. Today they only allowed me only three bounty $\endgroup$ – gymbvghjkgkjkhgfkl May 5 '17 at 17:18
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The correct prosthaphaeresis formula is $$ 2\sin{A}\cos{B} = \sin{(A+B)}+\sin{(A-B)}. $$ This gives $$ \frac{1}{2} \int_{-\infty}^{\infty} (\sin{(e^{-x})} + \sin{(e^{x})}) \, dx, $$ from the definitions of the hyperbolic functions, $$ \cosh{x}+\sinh{x} = e^x, \qquad \cosh{x}-\sinh{x} = e^{-x}. $$ The integrand is an even function, so this is the same as $$ \int_{-\infty}^{\infty} \sin{(e^{x})} \, dx $$ by putting $y=-x$ in one of the terms. Now put $y=e^{x}$. The limits change to $0$ and $\infty$, and $dx=dy/y$, so the integral becomes $$ \int_0^{\infty} \frac{\sin{y}}{y} \, dy, $$ which can be done in numerous ways.

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