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

Question

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

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!

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 07:52
@Christopher Carl Heckman:Thanks,Will you please tell me how to make the sign of definite integral. — Styles, Apr 28 '16 at 07:54
@ChristopherCarlHeckman:actually,I got a problem to find the inverse sine transformation of exp(-sk)/k — Styles, Apr 28 '16 at 07:56
Do you mean inverse hyperbolic sine transformation, by any chance? — Christopher Carl Heckman, Apr 28 '16 at 07:58
@ChristopherCarlHeckman:NO,its ordinary sine function with argument (k*x) — Styles, Apr 28 '16 at 08:00
@ChristopherCarlHeckman:Sorry! its Inverse fourier sine transformation. — Styles, Apr 28 '16 at 08:06
That's a new one for me. The answer is at http://dlmf.nist.gov/1.14#vii , but they don't say how they got it. ... This question was also answered at http://math.stackexchange.com/questions/878788/fourier-sine-transform-of-frac1x — Christopher Carl Heckman, Apr 28 '16 at 08:28

Olivier Oloa · Accepted Answer · 2016-04-28T08:25:20.937

4

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

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

edited Apr 28 '16 at 08:25

answered Apr 28 '16 at 08:08

Olivier Oloa

120,989

: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!! – Styles Apr 28 '16 at 08:18
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 08:20
:Thanks,I got it. – Styles Apr 28 '16 at 08:22
You are welcome. – Olivier Oloa Apr 28 '16 at 08:22
:Will you please tell me how to make the definite integral sign.I've already checked the MathsJax tutorial. – Styles Apr 28 '16 at 08:28
\int_a^b f(x),dx – Christopher Carl Heckman Apr 28 '16 at 08:37

Jean Marie · Answer 2 · 2016-04-28T09:29:39.323

0

@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.

edited Apr 28 '16 at 09:29

answered Apr 28 '16 at 09:07

Jean Marie

81,803

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 07:00
Not only to Laplace Transform but to the cardinal sine, two points that no other colleague had underlined. – Jean Marie Apr 30 '16 at 07:45

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

2 Answers2