How Does Integration by Parts Apply in Calculating Average Energy?

[Moneer81]

Hi,

while solving for the average energy given by the following formula:

$$\overline {E} = \frac {\int_{0}^{\infty} E e^\frac{-E}{kT}dE}{\int_{0}^{\infty} e^\frac{-E}{kT}dE}$$

where E bar is average energy, k is the Boltzmann's constant, and T is temperature

I had to use integration by parts for the numerator.

Integration by parts formula is $$\int u dv = uv - \int v du$$

So I made the following choices (and so did my textbook):

$$u = E$$

then $$du = dE$$

$$dv = e^\frac{-E}{kT}$$

and so $$v = -kTe^\frac{-E}{kT}$$

Then I proceeded by applying the integration by parts formula, and the integral of the numerator would be:

$$\int_{0}^{\infty} E e^\frac{-E}{kT}dE = -EkTe^\frac{-E}{kT} + kT \int_{0}^{\infty} e^\frac{-E}{kT}dE$$

but to my surprise, the book proceeded in the following manner:

$$\int_{0}^{\infty} E e^\frac{-E}{kT}dE = kT \left[e^\frac{-E}{kT} \right]<br /> _{0}^{\infty} + kT \int_{0}^{\infty} e^\frac{-E}{kT}dE$$

That first term to the right of the equal sign threw me off...where did it come from?

 

[dextercioby]

First

$$dv=e^{-\frac{E}{kT}} \ dE$$

then you're right. It has to be some error in the book. Perhaps they meant

$$kT \left[e^{-\frac{E}{kT}}\right]_{+\infty}^{0}$$

,that is converting the minus before the whole term into an interchange of integration limits.

Daniel.