Laplace transform of $\sqrt[3]{t}$ using Gamma function

Question

Can I ask what is the Laplace transform of $\sqrt[3]{t}$ using the Gamma function? This was my initial answer. Note that there is a theorem, $\Gamma(\frac{1}{3})=3\Gamma(\frac{4}{3})$. $$\int^{\infty}_{0}\exp^{-\beta} \beta^{-\frac{2}{3}}=3\int^{\infty}_{0}\exp^{-\beta} \beta^{\frac{4}{3}}$$ And was able to get the final answer of $$\Gamma(\frac{1}{3})=9\exp^{-\beta} \beta^{\frac{4}{3}}|^{\infty}_{0}+12\exp^{-\beta} \beta^{\frac{1}{3}}|^{\infty}_{0}$$ I don't know then whats the next thing to do.

Answer 1

Laplace of $t^n$ is $[\Gamma (n+1)]/[s^{(n+1)}]$

So if we put $n=1/2$, we get $\Gamma (1/2) ÷ (s^{1/2})$ and $\Gamma(1/2) = \sqrt{\pi}$.

Hence the answer is $\sqrt{\pi/s}$. Hope this will help calculating the Laplace of $t^n$.

Answer 2

The Gamma function can be defined as

$$\Gamma(z)=\int_0^\infty\text{d}x\,e^{-x}x^{z-1}$$

The Laplace transform you want to compute is

$$\mathcal{L}\left\{t^{1/3}\right\}=\int_0^\infty\text{d}t\,e^{-ts}t^{1/3}$$

Performing the change of variables $ts=y$ you have

$$\begin{align}\mathcal{L}\left\{t^{1/3}\right\}&=\int_0^\infty s^{-1}\text{d}y\,e^{-y}y^{1/3}s^{-1/3}\\&=s^{-4/3}\int_0^\infty \text{d}y\,e^{-y}y^{\frac{4}{3}-1}\\ &=s^{-4/3}\Gamma(4/3)\end{align}$$

Answer 3

The correct proof needs some complex analysis.

Define for $Re(s) > 0$ : $$\Gamma(s) = \int_0^\infty t^{s-1}e^{-t}dt$$

Then for real $s \in (0,\infty)$ you have $$\mathcal{L}[t^{1/3}1_{ t > 0}](s) = \int_0^\infty t^{1/3} e^{-st} dt \underset{x \ = \ st}= s^{-4/3}\int_0^\infty x^{1/3} e^{-x} dx = s^{-4/3}\Gamma(4/3)$$

Now note that both the LHS and the RHS are complex analytic for $Re(s) > 0$, and since they are equal for $s \in (0,\infty)$, by the identity theorem for complex analytic functions (or if you prefer by analytic continuation) it proves they are equal for every $Re(s) > 0$.

(the idea of the identity theorem is that if a function $h(s)$ is complex analytic and non identically zero, then its zeros are isolated, but here $\mathcal{L}[t^{1/3}1_{ t > 0}](s)-s^{-4/3}\Gamma(4/3)$ is complex analytic on $Re(s) > 0$ and identically zero on the real line, hence it is identically zero on $Re(s) > 0$)

Answer 4

$$\mathcal{L}_t\left[t^{\frac{1}{\text{n}}}\right]_{\text{s}}=\int_0^\infty t^{\frac{1}{\text{n}}}\cdot e^{-\text{s}t}\space\text{d}t=\text{s}^{-\frac{1+\text{n}}{\text{n}}}\Gamma\left\{1+\frac{1}{\text{n}}\right\}$$

When $\Re\left[\text{s}\right]>0\space\wedge\space\Re\left[\frac{1}{\text{n}}\right]>-1$