Laplace tranform of $t^{5/2}$

Question

It is asked to transform $t^{5/2}$.

I did $t^{5/2}=t^3\cdot t^{-1/2}$. Then followed the table result $$L\{{t^nf(t)}\}=(-1)^n\cdot\frac{d^n}{ds^n}F(s)$$

However i got $\frac{1}{2} \cdot\sqrt\pi \cdot s^{-7/2}$ instead of $\frac{15}{8} \cdot\sqrt\pi \cdot s^{-7/2}$.

Can you help me with the derivations?

Thanks

Answer 1

Using the result $$L\{{t^nf(t)}\}=(-1)^n \cdot\frac{d^n}{ds^n}F(s).$$

• $t^{5/2}=t^3\cdot t^{-1/2}$.

• $ L(t^{-1/2})=\sqrt{\large\frac{\pi}{s}}=s^{-1/2}\sqrt{\pi}$, now apply the result.

$$L\{{t^3\cdot t^{-1/2}}\}=(-1)^3\cdot\frac{d^3}{ds^3}\{s^{-1/2}\sqrt{\pi}\}=\frac{15}{8}\sqrt{\pi}\cdot s^{-\frac{7}{2}}.$$

Answer 2

For every real number $r>-1$, we have,

$$L(t^r)(s)=\int_0^\infty e^{-st}t^rdt\\\hspace{60mm}=\int_0^\infty e^{-x}(\frac{x}{s})^r\frac{dx}{s}\hspace{10mm}\text{(Putting $x=st$)}\\\hspace{20mm}=\frac{1}{s^{r+1}}\int_0^\infty e^{-x}x^rdt\\\hspace{5mm}=\frac{\Gamma(r+1)}{s^{r+1}}.$$

So for $r=\frac{5}{2}$, we have, $$L(t^{\frac{5}{2}})(s)=\frac{\Gamma(\frac{5}{2}+1)}{s^{\frac{5}{2}+1}}=\Gamma(\frac{7}{2})s^{-\frac{7}{2}}$$

Now we get $$\Gamma(\frac{7}{2})=\Gamma(\frac{5}{2}+1)=\frac{5}{2}\Gamma(\frac{5}{2})=\frac{5}{2}\Gamma(\frac{3}{2}+1)=\frac{5}{2}\frac{3}{2}\Gamma(\frac{1}{2}+1)\\=\frac{5}{2}\frac{3}{2}\frac{1}{2}\Gamma(\frac{1}{2})=\frac{15}{8}\sqrt{\pi}.$$

$$\therefore L( t^{\frac{5}{2}})(s)=\frac{15}{8}\sqrt{\pi}\cdot s^{-\frac{7}{2}}.$$

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For $p>0$ the Euler gamma function $\Gamma(p)$ is defined as, $$\Gamma(p)=\int_0^\infty e^{-x}x^{p-1}dx$$

A property of gamma function: $$\Gamma(p+1)=p\cdot\Gamma(p)$$

Also $\Gamma(\frac{1}{2})=\sqrt{\pi}$.