Inverse Laplace transform of $\frac{a\sinh\left[\frac{p}{c}(L-x)\right]}{p\sinh\left[\frac{pL}{c}\right]}$ with respect to $p$ using binomial theorem

Question

I have the expression $$\frac{a\sinh\left[\frac{p}{c}(L-x)\right]}{p\sinh\left[\frac{pL}{c}\right]}$$ and I want to find the inverse Laplace transform as an infinite series by using the binomial expansion. I tried rewriting the denominator as exponentials and expanding using the binomial theorem, but this gave me an infinite series of terms of the form $\exp\cdot\sinh$ which when inverse Laplace transformed gave me a delta function using the convolution theorem (which doesn't seem right - I don't think my answer should be an infinite sum of delta functions!).

How can I use the binomial theorem here?

@JeanMarie p is indeed the Laplace variable, and all other things are constants with respect to p. It is explicitly given that I should use the binomial theorem, although it is certainly possible I could combine it with Taylor expansions. — acernine, May 31 '21 at 00:02
@JeanMarie We can write $\operatorname {csch} p$ as a geometric series of exponentials (convergent for $\operatorname {Re} p > 0$), which of course is a special case of the binomial series. Since there is a $p$ factor in the denominator, we get a sum of shifted unit step functions, or $\mathcal L^{-1}p \mapsto p^{-1} \sinh(p + a) \operatorname {csch} p = A \lfloor t/2 \rfloor + B$ for some constants $A$ and $B$. — Maxim, Jun 02 '21 at 10:50
@Maxim I haven't had chance to go over the details, but it sounds like this is exactly what I'm looking for, thank you! — acernine, Jun 02 '21 at 11:13
@Maxim I don't understand. A geometric series isn't the same as a binomial expansion (think in particular to the factorials that aren't present in a geometic series). — Jean Marie, Jun 02 '21 at 12:19

Jean Marie · Answer 1 · 2022-07-09T22:51:08.677

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Here is an answer to the question but without any use of binomial theorem.

Let us use simplified notations:

$$F(s)=\frac{a}{p}\frac{\sinh(Ap-B)}{\sinh(Ap)}= \frac{a}{p}\left(\frac{\sinh(Ap)\cosh(B)-\sinh(B)\cosh(Ap)}{\sinh(Ap)}\right)$$

$$F(s)=\frac{a}{p}\cosh(B)-a\sinh(B)\frac{1}{p}\operatorname{coth}(Ap)\tag{1}$$

Here, we can use the (ill-known) series representation:

$$\operatorname{coth}(X)=\frac1X + 2\sum_{n=1}^{\infty} \frac{X}{\pi^2n^2+X^2}$$

giving

$$\frac{1}{p}\operatorname{coth}(Ap)=\frac{1}{Ap^2} + 2A \sum_{n=1}^{\infty} \frac{1}{\pi^2n^2+(Ap)^2}\tag{2}$$

From there it is not difficult to obtain the inverse Laplace Transform of (2) due to classical formula:

$$\mathcal{L}^{-1}\frac{1}{k^2+p^2}=\frac1k \sin kt$$

under the form of an infinite series, in fact the Fourier series of a sawtooth function.

But where is the Binomial Theorem hidden ?

edited Jul 09 '22 at 22:51

answered May 31 '21 at 09:05

Jean Marie

81,803

Any comment ?... I have in fact given you an answer even if some little work is left... but I doubt more and more that there is a method using the binomial theorem, even the extended binomial theorem under the form $(1+X)^{\alpha}=1+\alpha X+\alpha( \alpha-1)/2 X^2+...$ – Jean Marie May 31 '21 at 12:27
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Thanks for the answer, this is a nice solution although I'll leave the question open as I'm looking for a solution using the binomial theorem. – acernine Jun 02 '21 at 11:15
There's a factor $A$ missing in Eq. (2). I tried to edit this but apparently edits need to be more than 6 characters.. – adriaanJ Jul 07 '22 at 09:31
Thanks @adriaanvalk – Jean Marie Jul 07 '22 at 16:06
The edit is wrong, there's still a factor $A$ missing in the numerator of the last term of eq. (2) and the factor $p$ in denominator should not be there. – adriaanJ Jul 08 '22 at 10:16
@adriaanvalk Yes you are right ! Fixed now... – Jean Marie Jul 09 '22 at 22:49

Inverse Laplace transform of $\frac{a\sinh\left[\frac{p}{c}(L-x)\right]}{p\sinh\left[\frac{pL}{c}\right]}$ with respect to $p$ using binomial theorem

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