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Evaluate the integral $$f(t)=\int_0^{\pi/2}\sin(t\tan x)dx.$$

I came across this integral after having shown that its counterpart can be evaluated exactly: $$\int_0^{\pi/2}\cos(t\tan x)dt=\frac{\pi}{2}e^{-|t|}.$$ I figured that $f$ would be as easy but it is actually presenting me with some difficulties. I start off by using the substitution $\tan x\mapsto x$: $$f(t)=\int_0^\infty \frac{\sin tx}{1+x^2}dx,$$ then, taking the Laplace transform: $$\begin{align} \mathcal{L}\{f\}(s)&=\int_0^\infty \frac1{1+x^2}\int_0^\infty e^{-st}\sin(xt)dtdx\\ &=\int_0^\infty \frac1{1+x^2}\text{Im}\left[\int_0^\infty e^{-(s-ix)t}dt\right]dx\\ &=\int_0^\infty \frac1{1+x^2}\frac{x}{s^2+x^2}dx\\ &=\frac12\int_0^\infty \frac{dx}{(1+x)(s^2+x)}\\ &=\frac1{2s^2-2}\int_0^\infty \left[\frac{1}{1+x}-\frac{1}{s^2+x}\right]dx\\ &=\frac{\ln s}{s^2-1}. \end{align}$$ Now, I see that $f(t)=\mathcal{L}^{-1}\left\{\frac{\ln s}{s^2-1}\right\}(t),$ but I do not know how to compute that inverse Laplace transform. Could I have some help?

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  • $\begingroup$ $f(t)=\text{Shi}(t) \cosh t-\text{Chi}(t) \sinh t$, it seems to be non-elementary. $\endgroup$ – Kemono Chen Sep 2 at 4:59
  • $\begingroup$ @KemonoChen what are your $\text{Shi}$ and $\text{Chi}$ functions here? $\endgroup$ – clathratus Sep 2 at 5:01
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    $\begingroup$ Shi and Chi are hyperbolic sine/cosine integral respectively. $\endgroup$ – Kemono Chen Sep 2 at 5:02
  • $\begingroup$ Ahh I see... Thanks. $\endgroup$ – clathratus Sep 2 at 5:03
  • $\begingroup$ @KemonoChen. If I may ask, how did you arrive to this nice and simple result ? I am just curious. I suppose that intermediate steps involve some complex expressions. $\endgroup$ – Claude Leibovici Sep 2 at 5:17
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After Kemono Chen's comment.

Starting with $$f(t)=\int_0^\infty \frac{\sin (tx)}{1+x^2}\,dx=\int_0^\infty \frac{\sin (tx)}{(x+i)(x-i)}\,dx$$ $$f(t)=\frac i 2 \left(\int_0^\infty \frac{\sin (tx)}{x+i}\,dx -\int_0^\infty \frac{\sin (tx)}{x-i}\,dx\right)$$ Consider now $$I(a)=\int \frac{\sin (tx)}{x+a}\,dx=\int\frac{\sin (t (y-a))}{y}\,dy$$ $$I(a)=\cos(at) \int \frac{\sin (ty)}{y}\,dy-\sin(at)\int \frac{\cos (ty)}{y}\,dy$$ Now, make $z=ty$ $$I(a)=\cos(at) \int \frac{\sin (z)}{z}\,dz-\sin(at)\int \frac{\cos (z)}{z}\,dz=\cos (a t)\,\text{Si}(z) -\sin (a t)\,\text{Ci}(z) $$ Back to $x$ $$I(a)=\cos (a t)\, \text{Si}(t(x+a))-\sin (a t)\, \text{Ci}(t(x+a))$$ Assuming $t >0$ $$J(a)=\int_0^\infty \frac{\sin (tx)}{x+a}\,dx=\text{Ci}(a t) \sin (a t)+\frac{1}{2} (\pi -2 \text{Si}(a t)) \cos (a t)$$ $$J(a)-J(-a)=(\text{Ci}(-a t)+\text{Ci}(a t)) \sin (a t)-2 \text{Si}(a t) \cos (a t)$$ $$f(t)=\frac i 2 \left(J(i)-J(-i)\right)=\text{Shi}(t) \cosh (t)-\text{Chi}(t) \sinh (t)$$ which is Kemono Chen's result.

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  • $\begingroup$ So, how to we get to Kemono Chen's result from here? $\endgroup$ – clathratus Sep 2 at 16:34
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    $\begingroup$ @clathratus. I shall elaborate tomorrow morning. It is dinner time here. Cheers $\endgroup$ – Claude Leibovici Sep 2 at 16:36
  • $\begingroup$ Very nice, thanks :) $\endgroup$ – clathratus Sep 3 at 20:55
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Mathematica gives:

$$\frac{1}{4} \sqrt{\pi } t G_{1,3}^{2,1}\left(\frac{t^2}{4}| \begin{array}{c} 0 \\ 0,0,-\frac{1}{2} \\ \end{array} \right)$$

involving the Meier G function.

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    $\begingroup$ @clathratus: Well, sure. But seeing the correct result might make you reconsider your wish to find a hand derivation. Good luck to you... you'll need it!! $\endgroup$ – David G. Stork Sep 2 at 5:28
  • $\begingroup$ Thank you for your considerations then, I hadn't thought of it that way (+1) $\endgroup$ – clathratus Sep 2 at 19:31

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