# Find the Laplace transform for this function f(x)=(1+2ax)x^(-1/2) e^(ax)

Find the Laplace transform for this function
$f\left(x\right)=\left(1+2ax\right){x}^{-\frac{1}{2}}{e}^{ax}$
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Remember the Laplace transform
$L\left[{e}^{at}{t}^{n}\right]=\frac{\mathrm{\Gamma }\left(n+1\right)}{\left(s-a{\right)}^{n+1}}$
Now, we have
$L\left[\left(1+2ax\right){x}^{-1/2}{e}^{ax}\right]$
$=L\left[{e}^{ax}{x}^{-1/2}\right]+2aL\left[{e}^{ax}{x}^{1/2}\right]$
$=\frac{\mathrm{\Gamma }\left(-\frac{1}{2}+1\right)}{\left(s-a{\right)}^{-\frac{1}{2}+1}}+2a\frac{\mathrm{\Gamma }\left(\frac{1}{2}+1\right)}{\left(s-a{\right)}^{\frac{1}{2}+1}}$
$=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\sqrt{s-a}}+2a\frac{\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\left(s-a{\right)}^{3/2}}$
$=\frac{\sqrt{\pi }}{\sqrt{s-a}}+a\frac{\sqrt{\pi }}{\left(s-a{\right)}^{3/2}}$
$=\frac{s\sqrt{\pi }}{\left(s-a{\right)}^{3/2}}$
###### Did you like this example?
propappeale00
$f\left(x\right)=\left(1+2ax\right){x}^{-\frac{1}{2}}{e}^{ax}$
$f\left(x\right)={x}^{-\frac{1}{2}}{e}^{ax}+2a{x}^{\frac{1}{2}}{e}^{ax}$
$\overline{f}\left(s\right)=L\left[{x}^{-\frac{1}{2}};s-a\right]+2aL\left[{x}^{\frac{1}{2}};s-a\right]$
But
$L\left[{x}^{-\frac{1}{2}};s\right]={\int }_{0}^{\mathrm{\infty }}{e}^{-sx}{x}^{-\frac{1}{2}}dx$
let $t=sx$ then $dt=s.dx$ , t from $0\to \mathrm{\infty }$
$=\frac{1}{\sqrt{s}}{\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{-\frac{1}{2}}dt$
$=\frac{1}{\sqrt{s}}{\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{1}{2}-1}dt=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\sqrt{s}}=\sqrt{\frac{\pi }{s}}$
So
$L\left[{x}^{-\frac{1}{2}};s-a\right]=\sqrt{\frac{\pi }{s-a}}$
And
$L\left[{x}^{\frac{1}{2}};s\right]={\int }_{0}^{\mathrm{\infty }}{e}^{-sx}{x}^{\frac{1}{2}}dx$
let $t=sx$ then $dt=s.dx$ , t from $0\to \mathrm{\infty }$
$=\frac{1}{\sqrt{s}}{\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{1}{2}}dt$
$=\frac{1}{{s}^{\frac{3}{2}}}{\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{3}{2}-1}dt=\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)}{{s}^{\frac{3}{2}}}=\frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{{s}^{\frac{3}{2}}}=\frac{1}{2s}\sqrt{\frac{\pi }{s}}$
So
$L\left[{x}^{\frac{1}{2}};s-a\right]=\frac{1}{2\left(s-a\right)}\sqrt{\frac{\pi }{s-a}}$
Thus
$\overline{f}\left(s\right)=\sqrt{\frac{\pi }{s-a}}+2a\frac{1}{2\left(s-a\right)}\sqrt{\frac{\pi }{s-a}}$
$\overline{f}\left(s\right)=\sqrt{\frac{\pi }{s-a}}+\frac{a}{\left(s-a\right)}\sqrt{\frac{\pi }{s-a}}$
$\overline{f}\left(s\right)=\left[1+\frac{a}{\left(s-a\right)}\right]\sqrt{\frac{\pi }{s-a}}$
$\overline{f}\left(s\right)=\left[\frac{s}{\left(s-a\right)}\right]\sqrt{\frac{\pi }{s-a}}$
$\overline{f}\left(s\right)=\frac{s\sqrt{\pi }}{\left(s-a{\right)}^{\frac{3}{2}}}$