# Find the Laplace transform for this function

Find the Laplace transform for this function

$$f(x)=(1+2ax)x^{-\frac{1}{2}}e^{ax}$$

Remember the Laplace transform $$L[e^{at}t^n]=\frac{\Gamma{(n+1)}}{(s-a)^{n+1}}$$

Now, we have $$L[(1+2ax)x^{-1/2}e^{ax}]$$ $$=L[e^{ax}x^{-1/2}]+2aL[e^{ax}x^{1/2}]$$ $$=\frac{\Gamma{\left(-\frac{1}{2}+1\right)}}{(s-a)^{\large -\frac12+1}}+2a\frac{\Gamma{\left(\frac{1}{2}+1\right)}}{(s-a)^{\large \frac12+1}}$$ $$=\frac{\Gamma{\left(\frac{1}{2}\right)}}{\sqrt{s-a}}+2a\frac{\frac{1}{2}\Gamma{\left(\frac{1}{2}\right)}}{(s-a)^{3/2}}$$ $$=\frac{\sqrt \pi}{\sqrt{s-a}}+a\frac{\sqrt\pi}{(s-a)^{3/2}}$$ $$=\color{red}{\frac{s\sqrt \pi}{(s-a)^{3/2}}}$$

My Solution:- $$f(x)=(1+2ax)x^{-\frac{1}{2}}e^{ax}$$ $$f(x)=x^{-\frac{1}{2}}e^{ax}+2ax^{\frac{1}{2}}e^{ax}$$ $$\bar f(s)=L[x^{-\frac{1}{2}};s-a]+2aL[x^{\frac{1}{2}};s-a]$$

But $$L[x^{-\frac{1}{2}};s]=\int_{0}^{\infty}e^{-sx}x^{-\frac{1}{2}} dx$$ let $$t=sx$$ then $$dt=s.dx$$ , $$t$$ from $$0 \to \infty$$ $$= \frac{1}{\sqrt{s}}\int_{0}^{\infty}e^{-t}t^{-\frac{1}{2}} dt$$ $$= \frac{1}{\sqrt{s}}\int_{0}^{\infty}e^{-t}t^{\frac{1}{2}-1} dt =\frac{\Gamma(\frac{1}{2})}{\sqrt{s}}=\sqrt{\frac{\pi}{s}}$$

So $$L[x^{-\frac{1}{2}};s-a]=\sqrt{\frac{\pi}{s-a}}$$

And $$L[x^{\frac{1}{2}};s]=\int_{0}^{\infty}e^{-sx}x^{\frac{1}{2}} dx$$ let $$t=sx$$ then $$dt=s.dx$$ , $$t$$ from $$0 \to \infty$$ $$= \frac{1}{\sqrt{s}}\int_{0}^{\infty}e^{-t}t^{\frac{1}{2}} dt$$ $$= \frac{1}{s^\frac{3}{2}}\int_{0}^{\infty}e^{-t}t^{\frac{3}{2}-1} dt =\frac{\Gamma(\frac{3}{2})}{s^\frac{3}{2}}=\frac{1}{2}\frac{\Gamma(\frac{1}{2})}{s^\frac{3}{2}}=\frac{1}{2s}\sqrt{\frac{\pi}{s}}$$

So $$L[x^{\frac{1}{2}};s-a]=\frac{1}{2(s-a)}\sqrt{\frac{\pi}{s-a}}$$

Thus $$\bar f(s)=\sqrt{\frac{\pi}{s-a}}+2a\frac{1}{2(s-a)}\sqrt{\frac{\pi}{s-a}}$$ $$\bar f(s)=\sqrt{\frac{\pi}{s-a}}+\frac{a}{(s-a)}\sqrt{\frac{\pi}{s-a}}$$ $$\bar f(s)=[1+\frac{a}{(s-a)}]\sqrt{\frac{\pi}{s-a}}$$ $$\bar f(s)=[\frac{s}{(s-a)}]\sqrt{\frac{\pi}{s-a}}$$ $$\bar f(s)=\frac{s\sqrt{\pi}}{(s-a)^\frac{3}{2}}$$

Is this true solution?

Is there a simplification of the final answer