Problems Integrating an Infinite Series From E&M

JKreutz
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Hi, guys. For an E&M/quantum mechanics problem I have to integrate the series below:

Homework Statement



Integrate

[itex]\int{r^{2}e^{y r} \, dr}.[/itex]2. The attempt at a solution

Using integration by parts and and "differentiating under the integral" give the same answer:

[itex]I= \frac{2 e^{yr}}{y^{3}}-\frac{2 r e^{yr}}{y^{2}}+\frac{r^{2} e^{yr}}{y} + C[/itex],

where I is the solution to the integral and C is an arbitrary constant. I'd read about Newton trying to integrate infinite series, so I thought as a check and out of curiosity I'd express the integrand as an infinite series and integrate each term. Specifically I expressed exp(yr) as a Maclauren series, multiplied each term by the [itex]r^{2}[/itex] and used the "power rule" to get

[itex]I=\sum_{n=0}^{\infty} {\frac{y^{n} r^{(n+2)}}{n! \, (n+3)}}.[/itex]

I can't see how this series could represent the other answer I found. Moreover, I've never seen this done before and I don't know to look up this kind of information. (Googling "Infinite Series and Integrals" was not helpful ><). Any help would be appreciated.
 
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I wonder if you did it right, if you set r=0 then you should get the answer y^-3, but looking at the series, I don't think you do.
 
JKreutz said:
Specifically I expressed exp(yr) as a Maclauren series, multiplied each term by the [itex]r^{2}[/itex] and used the "power rule" to get

[itex]I=\sum_{n=0}^{\infty} {\frac{y^{n} r^{(n+2)}}{n! \, (n+3)}}.[/itex]

There is a mistake, the power of r has to be n+3. And you need to add a constant to the power series integral, too. Replace exp(yr) in the first integral by its power series. Supposed you are in the range of convergence, the power-series solution is equivalent to the first integral if you match the constants. Find the coefficients of r, r2, r3 and so on and compare them.

ehild
 

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