Integral of squareroot and exponetial

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Homework Statement


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


Homework Equations





The Attempt at a Solution



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

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:

<br /> \int u^2\,{e}^{-bu^2}=\int u\cdot u{e}^{-bu^2}<br />

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

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

thanks
i was able to do it :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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