# Evaluate the improper integral $\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk$.

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

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 $\displaystyle\frac1k$.

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

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

• :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!! – P.Styles Apr 28 '16 at 8:18
• For $s\leq0$ your initial integral does not exist. the case $x>0$ is just to avoid handling absolute value. Thanks. – Olivier Oloa Apr 28 '16 at 8:20
• :Thanks,I got it. – P.Styles Apr 28 '16 at 8:22
• You are welcome. – Olivier Oloa Apr 28 '16 at 8:22
• :Will you please tell me how to make the definite integral sign.I've already checked the MathsJax tutorial. – 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.

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