Is This Calculation of Work Involving an Exponential Function Correct?

[kabailey]

W=∫₀^∞PdV



P=e^-v2



Do I simply substitue P in the original equation, differentiate, then integrate?

W=∫₀^∞e^-v2dV

W=∫₀^∞(e^-v2)/2v

So far is this correct?

 

[LCKurtz]

W=∫₀^∞PdV
P=e^-v2
Do I simply substitue P in the original equation, differentiate, then integrate?

W=∫₀^∞e^-v2dV

W=∫₀^∞(e^-v2)/2v

So far is this correct?

No, it isn't. That integral is usually worked by a trick of multiplying it by itself and changing it to polar coordinates. See
http://en.wikipedia.org/wiki/Gaussian_integral
There they do it from $-\infty$ to $\infty$, but the same idea applies.

 

[kabailey]

No, it isn't. That integral is usually worked by a trick of multiplying it by itself and changing it to polar coordinates. See
http://en.wikipedia.org/wiki/Gaussian_integral
There they do it from $-\infty$ to $\infty$, but the same idea applies.

I was not familiar with the Gaussian intergral. Well since it's only half the distance I would divide by 2 ∴

W=∫₀^∞e^-v2dV

=(√∏)/2|^∞₀

so does that mean that the interval of ∞ to 0 is negligible?

 

[LCKurtz]

I was not familiar with the Gaussian intergral. Well since it's only half the distance I would divide by 2 ∴

W=∫₀^∞e^-v2dV

=(√∏)/2|^∞₀

so does that mean that the interval of ∞ to 0 is negligible?

The integrand is an even function so, yes, you can just divide the full integral by 2. The integal from on $[-\infty,0]$ equals the integral over $[0,\infty]$.

 

[kabailey]

thank you for your help. my final answer is W=(√∏)/2

 

[LCKurtz]

thank you for your help. my final answer is W=(√∏)/2

That's correct.