Question on integration by parts

[trmcclain]

I want to start out by saying that this is not a homework problem, this is something I'm trying to figure out for thesis work. If that should go in the homework problem section, I will gladly post there.

A certain mass model I'm working with (Bissant & Gerhard) has a particularly gross form.
exp(-(r'/rc)^2)*(1+r'/r0)^-α

Separating out constants and rearranging, I can massage it into this form:
λ^n*exp(-x^2)*(λ+x)^-n , where λ is constant and n is known (1.8 for those who want to know)

My problem is now I have to integrate this sucker, and I've never been particularly good at parts. I did find a guide and I was able to walk myself through it, however, my issue come here.
I separated it out as:
u=(λ+x)^-n, du=-n(λ+x)^(-n-1)dx
dv=exp(-x^2), v= <-- PROBLEM

Unless I integrate dv from 0 to infinity, the gaussian is sqrt(pi)/2 times an error function, which is useless to me.
So... Can I integrate the dv from 0 to infinity and set different limits to the subsequent integrals? i.e., can I have dv=exp(-x^2), v=sqrt(pi)/2, but when it comes to the uv-∫vdu, can I set that integral from 0 to some finite limit?

Any help is much appreciated, and as I said before, if this is better suited for another board, I will gladly post there.

 

[SteamKing]

Some integrals do not have closed form solutions in terms of other elementary functions and must be evaluated numerically. The error function is one such integral which is encountered quite frequently, especially in statistical work. There are many other such integrals, but the fact that these integrals do not have closed form solutions has not prevented anyone from using them. Elliptic functions are just one such case, just like the error function.

 

[Stephen Tashi]

So... Can I integrate the dv from 0 to infinity and set different limits to the subsequent integrals?

The procedure of integration by parts involves anti-derivatives, not definite integrals. So you can't evaluate any of the "[itex] \int [/itex] " involved between definite limits. You can express an anti-derivative as an integral where the upper limit of integration is a variable, such as [itex] \int cos(x) = \int_0^x {cos(w)} dw [/itex].

There might be a procedure "something like" integration by parts that uses definite integrals. Offhand, I don't remember it.

 

[Stephen Tashi]

On this forum, you're more likely to get answers to problems invovling complicated expressions if you write them in latex. (See https://www.physicsforums.com/showpost.php?p=3977517&postcount=3 )

I'm not sure exactly what integration needs to be done. As I make it out, you are asking for [itex] \int exp(-(r'/rc)^2)*(1+r'/r0)^{-α} dr [/itex].

Is [itex] r' [/itex] the derivative of a function [itex] r [/itex]?

 

[lurflurf]

The procedure of integration by parts involves anti-derivatives, not definite integrals. So you can't evaluate any of the "[itex] \int [/itex] " involved between definite limits. You can express an anti-derivative as an integral where the upper limit of integration is a variable, such as [itex] \int cos(x) = \int_0^x {cos(w)} dw [/itex].

There might be a procedure "something like" integration by parts that uses definite integrals. Offhand, I don't remember it.

anti-derivatives and definite integrals are related by the fundamental theorem of calculus

$$\left. \int_a^b \! u \, \mathop{dv}= u \, v \right| _a^b -\int_a^b \! v \, \mathop{du}$$

of course there are continuity assumptions