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Hi I am trying to prove this interesting integral $$ \mathcal{I}:=\int_0^\infty \frac{\sin a x-\sin b x}{\cosh \beta x}\frac{dx}{x}=2\arctan\left(\frac{\exp(\frac{a\pi}{2\beta})-\exp(\frac{b\pi}{2\beta})}{1+\exp{\big(\frac{(a+b)\pi}{2\beta}}\big)}\right), \qquad Re(\beta)>0. $$ This is related to another integral posted except that had cosines in the numerator instead of sine functions. The result is quite different, very interesting integrals. It sure looks like a Frullani but this one's result is quite different looking than just a logarithm function..I

A Frullani integral is $$ \int_0^\infty \frac{f(ax)-f(bx)}{x}dx=\big[f(0)-f(\infty)\big]\log \frac{b}{a}. $$ In this example, $\mathcal{I}$ has a term $\cosh \beta x$ in the denominator which it makes it appear different. Partial integration and splitting the integral up didn't work and ran into convergent issues. How can we solve this integral? Thank you.

Note $$ 2\cosh x=e^x+e^{-x}. $$

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  • $\begingroup$ Isn't there s formula for $\sin u\pm\sin v$ ? $\endgroup$ – Lucian May 22 '14 at 6:02
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Consider the integral \begin{align} \int_{b}^{a} \cos(\mu x) \ d\mu = \left[ \frac{\sin(x \mu)}{x} \right]_{b}^{a} = \frac{\sin(ax) - \sin(bx)}{x}. \end{align} This integral wil be used in the calculation of the integral in question. The integral to calculate is \begin{align} I = \int_{0}^{\infty} \frac{\sin(ax) - \sin(bx)}{x \ \cosh(\beta x)} \ dx \end{align} and can be seen to be \begin{align} I &= \int_{0}^{\infty} \frac{\sin(ax) - \sin(bx)}{x \ \cosh(\beta x)} \ dx \\ &= \int_{b}^{a} \left( \int_{0}^{\infty} \frac{\cos(\mu x)}{\cosh(\beta x)} \ dx \right) \ d\mu. \end{align} The inner integral can be quickly calculated by using the known integral \begin{align} \int_{0}^{\infty} \frac{\cosh(\alpha x)}{\cosh(\beta x)} \ dx = \frac{\pi}{2 \beta} \ \sec\left( \frac{\alpha \pi}{2 \beta} \right). \end{align} By letting $\alpha = i \mu$ this becomes \begin{align} \int_{0}^{\infty} \frac{\cos(\mu x)}{\cosh(\beta x)} \ dx = \frac{\pi}{2 \beta} \ sech\left( \frac{\mu \pi}{2 \beta} \right) \end{align} and leads to \begin{align} I &= \frac{\pi}{2 \beta} \ \int_{b}^{a} sech\left( \frac{\pi \mu}{2 \beta} \right) \ d\mu \\ &= \int_{(b\pi/2\beta)}^{(a\pi/2\beta)} sech(x) \ dx \\ &= 2 \left[ \tan^{-1}\left(e^{\frac{a\pi}{2\beta}} \right) - \tan^{-1}\left(e^{\frac{b\pi}{2\beta}} \right) \right] \end{align} which can be restated as \begin{align} \int_{0}^{\infty} \frac{\sin(ax) - \sin(bx)}{x \ \cosh(\beta x)} \ dx = 2 \tan^{-1}\left( \frac{e^{a\pi/2\beta} - e^{b\pi/2\beta}}{1 + e^{(a+b)\pi/2\beta}}\right). \end{align}

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  • $\begingroup$ Hi Leucippus! That result is not known to me, can you please share a link where it is proved? Many thanks! :) $\endgroup$ – Pranav Arora May 22 '14 at 19:25
  • $\begingroup$ @Pranav The integral is in the book $Tables of integrals, series, and products", 7th ed, found by the following link. It is formula 3.511.4, book page 371. Formula 3.511.2 is used in another problem proposed by Integrals.: f3.tiera.ru/2/M_Mathematics/MRef_References/… $\endgroup$ – Leucippus May 22 '14 at 20:09
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First do a substitution $A = a/\beta$, $B = b/\beta$ to reduce the integral to $$ \int_0^{\infty} \frac{\sin A x}{x \cosh x} dx -\int_0^{\infty} \frac{\sin B x}{x \cosh x} dx $$ These integrals were calculated in a previous question, and the result is

$$ \left(2 \arctan e^{A\pi/2} -\frac{\pi}{2}\right)- \left(2\arctan e^{B\pi/2} -\frac{\pi}{2}\right)= 2 \left( \arctan e^{A\pi/2} - \arctan e^{B\pi/2} \right) $$

By the tangent addition formula, we have $\tan (a-b)=\frac{\tan a - \tan b}{1+\tan a \tan b} $, hence $$\arctan a -\arctan b = \arctan \frac{a-b}{1+ab} $$ and this gives the desired result. (By the way, I wonder why someone would intentionally write the answer in such a complicated form!)

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  • $\begingroup$ I think you are off by a factor of 4. The answer in my post is correct. Thanks $\endgroup$ – Jeff Faraci May 22 '14 at 16:40
  • $\begingroup$ I see my mistake; it is corrected now. $\endgroup$ – user111187 May 22 '14 at 16:45
  • $\begingroup$ Thanks +1. To answer your question, perhaps nobody had realized a simpler expression who tried this before you:) I agree your result is cleaned up very nicely. Thanks for the correction. $\endgroup$ – Jeff Faraci May 22 '14 at 17:19

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