Integral of squareroot and exponetial

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SUMMARY

The discussion focuses on integrating the function \(\int \sqrt{x - k} e^{-bx} \, dx\). The user successfully transforms the integral using the substitution \(x - k = u^2\), leading to the integral \(\int u^2 e^{-b(u^2 - k)} \, du\). By applying integration by parts, the user derives the solution, which includes the Gaussian integral \(\int e^{-bu^2} \, du\). The final result is expressed in terms of \(u\) and the Gaussian distribution.

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Mechdude
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Homework Statement


is there a general method of integrating this type of integral:
[tex]\int \sqrt{x -k} e^{-bx}[/tex]


Homework Equations





The Attempt at a Solution



[tex]x-k={u}^{2}[/tex]
[tex]dx=2udu[/tex]
[tex]\int u^2 \,{e}^{{-b\left( {u}^{2}-k\right) }}du[/tex]
[tex]\int {u}^{2}\,{e}^{{-b\,{u}^{2}}+{b\,k}}du[/tex]

and it seems worse than its starting point
 
Last edited:
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Actually it's not worse... You can finish the problem by integrating by parts:

[tex] \int u^2\,{e}^{-bu^2}=\int u\cdot u{e}^{-bu^2}[/tex]

I don't know what happened but as I type my message, the browser won't process more tex-code, so I put it in raw form, but I hope, you can understand from it what I'm saying. So:

=u\cdot \frac{{e}^{-bu^2}}{-2b}+\frac 1{2b}\,\int {e}^{-bu^2}

And it's done, because the last integral can be calculated explicitly, see the Gauss-distribution at wikipedia.
 
here's your code

[tex]=u\cdot \frac{{e}^{-bu^2}}{-2b}+\frac 1{2b}\,\int {e}^{-bu^2}[/tex]

thanks
i was able to do it :)
 

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