Interpretation of the Following Summation

Question

I wanted to invert the following Laplacian expression which was not possible with simple manipulation.

$$ \frac{s}{s^2 + \omega^2}\exp\bigg(-\sqrt{\frac{s}{\alpha}}x\bigg) $$

Thus, I used the Taylor Series on each part and simplified it as shown:

$$ \implies \frac{1}{s} \bigg( 1 + \frac{\omega^2}{s^2}\bigg)^{-1}\exp\bigg(-\sqrt{\frac{s}{\alpha}}x\bigg) \implies \frac{1}{s}\bigg( \sum_{k = 0}^{\infty} \bigg(\frac{-\omega^2}{s^2}\bigg)^k \bigg) \cdot \bigg( \sum_{n = 0}^{\infty} \frac{1}{n!}\bigg( -\sqrt{\frac{s}{\alpha}}x \bigg)^n \bigg) $$

$$ \sum_{k = 0}^{\infty} (-\omega^2)^k \sum_{n=0}^{\infty} \frac{1}{n!} \bigg( \frac{-x}{\sqrt{\alpha}}\bigg)^n \frac{1}{s^{2k - n/2 + 1}} $$

Now, I simply have to apply Inverse Laplace transform on only 1 term $1/s^a$ which is simply $t^{a-1}/\Gamma(a)$. Thus, I get

$$ \sum_{k = 0}^{\infty} (-\omega^2)^k \sum_{n=0}^{\infty} \frac{1}{n!} \bigg( \frac{-x}{\sqrt{\alpha}}\bigg)^n \frac{t^{2k - n/2}}{\Gamma(2k - n/2 + 1)} $$

This can be re-arranged into

$$ \implies \boxed{\sum_{k = 0}^{\infty} (-\omega^2 t^2)^k \sum_{n=0}^{\infty} \frac{1}{n!} \bigg( \frac{-x}{\sqrt{\alpha t}}\bigg)^n \frac{1}{\Gamma(2k - n/2 + 1)}} $$

Now the issue is about how to interpret this summation. This is not a double summation but just a product of 2 different summations. And even bigger issue is about how to handle values where n > 2(2k + 1).

How do I handle this expression, do I take a particular k and then sum the whole expression till n is a large expression???

Answer 1

The correct formula for the double sum is $$\sum _{k=0}^{\infty } \color{red}{(-1)^k} (\omega t)^{2 k} \sum _{n=0}^{\infty } \frac{\left(-\frac{x}{\sqrt{\alpha t}}\right)^n}{n! \Gamma \left(2 k-\frac{n}{2}+1\right)}$$

The inner infinite sum evaluates to

$$\sum _{n=0}^{\infty } \frac{\left(-\frac{x}{\sqrt{\alpha t}}\right)^n}{n! \Gamma \left(2 k-\frac{n}{2}+1\right)}=\frac{\, _1F_1\left(-2 k;\frac{1}{2};-\frac{x^2}{4 t \alpha }\right)}{\Gamma (2 k+1)}-\frac{x \sqrt{\alpha t} \, _1F_1\left(\frac{1}{2}-2 k;\frac{3}{2};-\frac{x^2}{4 t \alpha }\right)}{\alpha t \Gamma \left(\frac{1}{2} (4 k+1)\right)}$$

Finally we get

$$\sum _{k=0}^{\infty } (-1)^k (\omega t)^{2 k} \left(\frac{\, _1F_1\left(-2 k;\frac{1}{2};-\frac{x^2}{4 t \alpha }\right)}{\Gamma (2 k+1)}-\frac{x \sqrt{\alpha t} \, _1F_1\left(\frac{1}{2}-2 k;\frac{3}{2};-\frac{x^2}{4 t \alpha }\right)}{\alpha t \Gamma \left(\frac{1}{2} (4 k+1)\right)}\right)$$

HINT: $_1F_1(a;b;z)$ is the Kummer confluent hypergeometric function of the first kind.

Visualize sum expression with $\alpha=1, \omega=2\pi, 0\le k \le 20$:

Answer 2

Using only Mathematica with The Convolution Theorem:

$$\mathcal{L}_s^{-1}\left[\frac{s}{s^2+\omega ^2}\right](t)=\cos (t \omega )$$

and

$$\mathcal{L}_s^{-1}\left[e^{-x \sqrt{\frac{s}{\alpha }}}\right](t)=\frac{e^{-\frac{x^2}{4 t \alpha }} x \alpha }{2 \sqrt{\pi } \sqrt{t^3 \alpha ^3}}$$

then:

$$\int_0^t \frac{e^{-\frac{x^2}{4 \alpha \tau }} x \alpha \cos ((t-\tau ) \omega )}{2 \sqrt{\pi } \sqrt{\alpha ^3 \tau ^3}} \, d\tau$$

This integral Mathematica can compute.

$\mathcal{L}_s^{-1}\left[\frac{e^{-x \sqrt{\frac{s}{\alpha }}} s}{s^2+\omega ^2}\right](t)=\\\frac{1}{4} e^{-i t \omega -x \sqrt{\frac{i \omega }{\alpha }}} \left(e^{x \left(\sqrt{-\frac{i \omega }{\alpha }}+\sqrt{\frac{i \omega }{\alpha }}\right)} \text{erfc}\left(\frac{x+2 t \sqrt{-i \alpha \omega }}{2 \sqrt{t \alpha }}\right)+e^{2 i t \omega } \left(\text{erfc}\left(\frac{x-2 t \sqrt{i \alpha \omega }}{2 \sqrt{t \alpha }}\right)+e^{2 x \sqrt{\frac{i \omega }{\alpha }}} \text{erfc}\left(\frac{x+2 t \sqrt{i \alpha \omega }}{2 \sqrt{t \alpha }}\right)\right)+e^{i \sqrt{2} x \sqrt{\frac{\omega }{\alpha }}} \text{erfc}\left(\frac{x+2 (-1)^{3/4} t \sqrt{\alpha \omega }}{2 \sqrt{t \alpha }}\right)\right)$

Code:

1/4 E^(-I t \[Omega] - x Sqrt[(I \[Omega])/\[Alpha]]) (E^( x (Sqrt[-((I \[Omega])/\[Alpha])] + Sqrt[(I \[Omega])/\[Alpha]])) Erfc[(x + 2 t Sqrt[-I \[Alpha] \[Omega]])/( 2 Sqrt[t \[Alpha]])] + E^(2 I t \[Omega]) (Erfc[(x - 2 t Sqrt[I \[Alpha] \[Omega]])/( 2 Sqrt[t \[Alpha]])] + E^(2 x Sqrt[(I \[Omega])/\[Alpha]]) Erfc[(x + 2 t Sqrt[I \[Alpha] \[Omega]])/( 2 Sqrt[t \[Alpha]])]) + E^(I Sqrt[2] x Sqrt[\[Omega]/\[Alpha]]) Erfc[(x + 2 (-1)^(3/4) t Sqrt[\[Alpha] \[Omega]])/( 2 Sqrt[t \[Alpha]])])