$$\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk$$

I've tried hard for this but of no use.I've applied integration by parts by which I get $$\int_0^\infty \exp(-sk)\sin(kx)\,dk=\frac{x}{x^2+ s^2}.$$ But, I'm not getting how to adjust $\displaystyle\frac1k$.

Please Help!

  • $\begingroup$ You need elliptic functions to do your integral. Are you sure you copied the problem correctly? Was this integral derived from something? $\endgroup$ – Christopher Carl Heckman Apr 28 '16 at 7:52
  • $\begingroup$ @Christopher Carl Heckman:Thanks,Will you please tell me how to make the sign of definite integral. $\endgroup$ – P.Styles Apr 28 '16 at 7:54
  • $\begingroup$ @ChristopherCarlHeckman:actually,I got a problem to find the inverse sine transformation of exp(-sk)/k $\endgroup$ – P.Styles Apr 28 '16 at 7:56
  • $\begingroup$ Do you mean inverse hyperbolic sine transformation, by any chance? $\endgroup$ – Christopher Carl Heckman Apr 28 '16 at 7:58
  • $\begingroup$ @ChristopherCarlHeckman:NO,its ordinary sine function with argument (k*x) $\endgroup$ – P.Styles Apr 28 '16 at 8:00

We assume $x>0$ and $s>0$.

Then by differentiating the following identity with respect to $s$, $$ f(s)=\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk $$ one may write $$ f'(s)=-\int_0^\infty \exp(−sk)\sin(kx)\,dk=-\frac{x}{x^2+ s^2} $$ giving $$ f(s)=-\arctan \left( \frac{s}x\right)+C. $$ Observing that, as $s \to \infty$, $f(s) \to 0$, we then obtain $C=\dfrac\pi2$. Thus

$$ \int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk=\frac\pi2-\arctan \left( \frac{s}x\right), \qquad s>0,\,x>0. $$

  • $\begingroup$ :I'm not gettnig why you've assumed x>0 and s>0.How will the question get affected if we ignore these assumptions.Thank you!! $\endgroup$ – P.Styles Apr 28 '16 at 8:18
  • $\begingroup$ For $s\leq0$ your initial integral does not exist. the case $x>0$ is just to avoid handling absolute value. Thanks. $\endgroup$ – Olivier Oloa Apr 28 '16 at 8:20
  • $\begingroup$ :Thanks,I got it. $\endgroup$ – P.Styles Apr 28 '16 at 8:22
  • $\begingroup$ You are welcome. $\endgroup$ – Olivier Oloa Apr 28 '16 at 8:22
  • $\begingroup$ :Will you please tell me how to make the definite integral sign.I've already checked the MathsJax tutorial. $\endgroup$ – P.Styles Apr 28 '16 at 8:28

@ChristopherCarlHeckman : it is a Laplace Transform (see remark at the end)

Precisely, it is the Laplace Transform (LT) of a very important function, the cardinal sine (sinc).

This LT can be found in most LT tables as

$$\int_0^\infty {\sin(k)\over k}\exp(−sk)\,dk=\frac\pi2-\arctan(s)$$

from which it is easy to deduce by an elementary change of variables:

$$\int_0^\infty {\sin(kx)\over k}\exp(−sk)\,dk=\frac\pi2-\arctan \left( \frac{s}x\right)$$

Remark 1: This result can also be written $arccot\left( \frac{s}x\right)$.

Remark 2: In the reference dlmf.nist.gov/1.14#vii given by ChristopherCarlHeckman the result they give is not the same... I don't understand.

  • $\begingroup$ I knew that it was related to a Laplace Transform. I just hadn't worked with it before. $\endgroup$ – Christopher Carl Heckman Apr 30 '16 at 7:00
  • $\begingroup$ Not only to Laplace Transform but to the cardinal sine, two points that no other colleague had underlined. $\endgroup$ – Jean Marie Apr 30 '16 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.