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According to Laplace transform tables

$\mathcal{L}\{f(t,k)\}=e^{-k\sqrt s}/\sqrt s $

has the solution

${f(t,k)=e^{-c^2}/\sqrt{t\pi}}$

where k must be a positive real value and

${c=k/\sqrt4t}$

Making use of the complex error function erf(z), we can directly calculate the indefinite integral

$\int {e^{st}}{e^{-k \sqrt s}/\sqrt s}ds =-i\sqrt{\pi}{erf(i\sqrt{st}-ic)}/\sqrt{t}$

which can easily be verified by differentiation of both sides with respect to s.

Dividing by 2i$\pi$ and putting in the integration range from s=-iR to s=+iR makes f(t,k) approach

${e^{-c^2}/\sqrt{t\pi}}Re(erf(M+i(M-c))$

when we let R and thus M =$\sqrt{Rt/2}$ approach infinity.

Now

${Re(erf(M+i(M-c))=erf(M)-2/\sqrt{\pi}}e^{-M^2}{\int_{0}^{M-c}e^{u^2}sin(2Mu)du}$

This indicates that the last integral approaches zero leaving only erf(M), which goes to 1 at infinity thus confirming the transform tables expression for f(t,k) when k > 0.

However changing the sign of k gives

$\mathcal{L}\{f(t,-k)\}=e^{k\sqrt s}/\sqrt s $

and one would expect to get

${f(t,-k)=e^{-(-c)^2}/\sqrt{t\pi}=e^{-c^2}/\sqrt{t\pi}}$

which of course implies f(t,-k) = f(t,k).

But if the transform tables insist that k must be positive, then this can not be true so I need help as to why just changing the sign of c makes the integral

$e^{-M^2}{\int_{0}^{M+c}e^{u^2}sin(2Mu)du}$

seem to have a problem going to zero at infinity considering the highly oscillatory effect of sin(2Mu).

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  • $\begingroup$ This question is closely related to my previous question "Elusive Laplace transform", which was down voted (maybe not detailed enough or maybe because I didn't use MatJax). Hope to have better luck this time. $\endgroup$
    – Jens
    Mar 23, 2018 at 15:11
  • $\begingroup$ It seems to me that if you replace k by $|k|$ then everything is fine. This does not change the original result and preserves the evenness of the LT as a function of k. $\endgroup$
    – Ian
    Mar 23, 2018 at 15:20

2 Answers 2

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The question concerns the limit of

$e^{-M^2}{\int_{0}^{M+c}e^{u^2}sin(2Mu)du}$

I tried many tricks like splitting the integral into infinitely small parts and summing the contributions hoping that the positive half of the sine would nearly be cancelled by the negative part but this could not completely counter the exponential effect.

Wondering why I didn't think of it before, I finally settled on the following explanation:

Split the range 0 to M+c into the two ranges 0 to M and M to M+c giving

$e^{-M^2}{\int_{0}^{M}e^{u^2}sin(2Mu)du}$+$e^{-M^2}{\int_{M}^{M+c}e^{u^2}sin(2Mu)du}$

According to the mean value theorem the first integral equals $M{e^{m^2-M^2}sin(2Mm)}$ where m is some value of u in the range $0 < m < M$. Hence the exponent is negative and the integral vanishes when M becomes infinite. (This corresponds to putting c = 0 so k = 0 and not just k > 0 seems to give convergence!)

In the second integral we substitute ${v = 2Mu-M^2}$ to get

${e^{-M^2}{\int_{M}^{M+c}e^{u^2}sin(2Mu)du=}(1/2M){\int_{0}^{2Mc}e^{v^2/{4M^2}+v}sin(2M^2 + v)dv}}$.

The numerical value of the last expression is obviously greater than the numerical value of the much simpler integral

$(1/2M){\int_{0}^{2Mc}e^{v}sin(2M^2 + v)dv}$

which can be calculated directly to equal

$(1/4M)e^{2Mc}(sin{2Mc} -cos(4M^2)cos(2Mc))+(1/4M)cos(4M^2)$

The exponent in this expression is positive so it oscillates wildly between +/- infinity as M approaches infinity. Hence the limit is undefined when c > 0 and this seems to be the reason for the transform tables requiring a positive k.

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To sum up, $-\sqrt p$ approaches the directions $-1 \pm i$ on the line $\operatorname{Re} p = \gamma > 0$, while $\sqrt p$ approaches the directions $1 \pm i$. You get an exponential (with a square root in the exponent) divergence in the integral of $e^{\sqrt p + t p}/\sqrt p$.

Taking the principal value integral doesn't help either, because the integral from $\operatorname{Im} p = -A$ to $\operatorname{Im} p = A$ behaves like the diverging exponential factor times an oscillating factor.

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