laplace transform
I am having difficulty to figure out to use a Laplace Transform Table formula to verify a particular case.
The inverse Laplace transform of $$L^{-1}\bigg[\exp{\bigg\{-x\sqrt{(s+h)/k}\bigg\}}\bigg]$$ happens to be $$\frac{x\exp{[-ht-(x^2/4kt)]}}{2\sqrt{\pi kt^3}}$$
The conversion we need for such case from a Laplace transform table is $$\frac{a}{2\sqrt{\pi t^3}}\exp{(-a^2/4t)}=L^{-1}\bigg[\exp{(-a\sqrt{s})}\bigg]$$
Now, the question is how can I utilize the above formula?
$$I(s)=\frac{6}{Ls^2 + Rs + \frac{1}{C}}$$
Having fixed $R$, $C$ and $L$ (for example $R=6$, $C=1/5$, $L=1$), we have:
$$I(s) = \frac{6}{s^2+6s+5}$$
First of all, find the zeros of denominator:
$$s^2 + 4s + 1= 0 \Rightarrow s = -5 \vee s = -1$$
Then $s^2+4s+1= (s+5)(s+1)$.
We can write $$I(s) = \frac{a}{s+5} + \frac{b}{s+1} = \frac{a(s+1)+b(s+5)}{(s+1)(s+5)} = \frac{s(a+b) + (a + 5b)}{s^2 + 6s + 5} = \frac{6}{s^2+6s+5}$$
Then:
$$\left\{\begin{array}{lcl}a+b & = & 0\\a + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -b\\-b + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -\frac{3}{2}\\b & = & \frac{3}{2} \end{array}\right.$$
Then:
$$I(s) = -\frac{\frac{3}{2}}{s+5} + \frac{\frac{3}{2}}{s+1}$$
The antitransformation yield to:
$$i(t) = -\frac{3}{2}e^{-5t} + \frac{3}{2}e^{-t}$$
To compute this ILT of $U$, we need to use the definition:
$$u(x,t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} $$
where $B = e^{P}$ and $a = \sqrt{P} x/2$.
To evaluate this integral, we use Cauchy's theorem and define the following contour integral:
$$\oint_C dz \, B e^{-a \sqrt{P+4 z}} e^{z t} $$
where $C$ is the following contour:
where the large arc has radius $R$ and the small arc has radius $\epsilon$. Along this contour, the above integral is equal to
$$\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} + i R \int_{\pi/2-\arcsin{(c/R)}}^{\pi} d\theta \,e^{i \theta}\, B e^{-a \sqrt{P+4 R e^{i \theta}}} e^{R t e^{i \theta}}\\ + e^{i \pi} \int_R^{P/4+\epsilon} dx \, B e^{-i a \sqrt{4 x-P}} e^{-x t} + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \, B e^{-a \sqrt{P+4 \epsilon e^{i \phi}}} e^{\epsilon t e^{i \phi}}\\ + e^{-i \pi} \int_{P/4+\epsilon}^R dx \, B e^{+i a \sqrt{4 x-P}} e^{-x t}+ i R \int_{\pi}^{3 \pi/2+\arcsin{(c/R)}} d\theta \,e^{i \theta}\, B e^{-a \sqrt{P+4 R e^{i \theta}}} e^{R t e^{i \theta}}$$
By Cauchy's theorem, this contour integral is zero. Further, we consider the limits as $R \to \infty$ and $\epsilon \to 0$; in this limit, the second, fourth, and sixth integrals vanish. This means that we have, by Cauchy's theorem:
$$\int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} = -\int_{\infty}^{P/4} dx \, B e^{-i a \sqrt{4 x-P}} e^{-x t} + \int_{P/4}^{\infty} dx \, B e^{+i a \sqrt{4 x-P}} e^{-x t} $$
or, the ILT is
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} &= \frac{B}{\pi} \int_{P/4}^{\infty} dx \, \sin{\left (a \sqrt{4 x-P} \right )} e^{-x t} \\ &= \frac{B}{\pi} e^{-P t/4} \int_0^{\infty} dy \,\sin{\left ( 2 a \sqrt{y} \right )} e^{-y t} \\ &= \frac{2 B}{\pi} e^{-P t/4} \int_0^{\infty} du \, u \, \sin{(2 a u)} e^{-t u^2} \\ &= \frac{2 B}{\pi} e^{-P t/4} \int_0^{\infty} du \, u^2 \frac{\sin{(2 a u)}}{u} e^{-t u^2} \\ &= -\frac{2 B}{\pi} e^{-P t/4} \frac{\partial}{\partial t} \int_0^{\infty} du \, \frac{\sin{(2 a u)}}{u} e^{-t u^2} \\ &= -\frac{B}{\pi} e^{-P t/4} \frac{\partial}{\partial t} \int_{-\infty}^{\infty} du \, \frac{\sin{v}}{v} e^{-(t/(4 a^2)) v^2}\end{align}$$
Now, to do the integral on the RHS we use Parseval's theorem for Fourier transforms.
$$\begin{align} \int_{-\infty}^{\infty} du \, \frac{\sin{v}}{v} e^{-(t/(4 a^2)) v^2} &= \frac1{2 \pi} 2 a \sqrt{\frac{\pi}{t}} \pi \int_{-1}^1 dk \, e^{-a^2 k^2/t} \\ &= 2 \sqrt{\pi} \int_0^{a/\sqrt{t}} dw \, e^{-w^2} \\ &= \pi \, \operatorname{erf}{\left (\frac{a}{\sqrt{t}} \right )}\end{align} $$
Hence,
$$\begin{align} \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} &= -\frac{B}{\pi} e^{-P t/4} \frac{\partial}{\partial t} \left [ \pi \, \operatorname{erf}{\left (\frac{a}{\sqrt{t}} \right )}\right ] \\ &= \frac{a B}{\sqrt{\pi t^3}} e^{-P t/4} e^{-a^2/t} \end{align}$$
and as $a = \sqrt{P} x/2$ and $B=e^P$, we have
$$u(x,t) = \frac{\sqrt{P} x}{2 \sqrt{\pi t^3}} e^P e^{-P t/4} e^{-P x^2/(4 t)} $$
Best Answer
Observe that if $\mathcal L\{f(t)\}=F(s)$, we have $\mathcal L\left\{\mathrm e^{-ht}f(t)\right\}=F(s+h)$ and $\mathcal L\{f(kt)\}=\frac{1}{k}F\left(\frac{s}{k}\right)$ for $k>0$ and then $$\mathcal L\left\{e^{-ht}f(kt)\right\}=\frac{1}{k}F\left(\frac{s+h}{k}\right)$$ so for $f(t)=\frac{a}{2\sqrt{\pi t^3}}\exp{(-a^2/4t)}$ with Laplcae transform $F(s)=\exp{(-a\sqrt{s})}$ we have $$ \mathcal L\left\{e^{-ht}f(kt)\right\}=\frac{1}{k}\exp\left(-a\sqrt{\frac{s+h}{k}}\right) $$ and then for $a=x$ we have \begin{align} \mathcal L^{-1}\left\{\exp\left(-x\sqrt{\frac{s+h}{k}}\right)\right\}&=e^{-ht}kf(kt)\\ &=e^{-ht}k\frac{x}{2\sqrt{\pi (kt)^3}}\exp{\left(-\frac{x^2}{4(kt)}\right)}\\ &=\frac{x}{2\sqrt{\pi kt^3}}\exp{\left(-ht-\frac{x^2}{4kt}\right)} \end{align}