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Kainui
Does anyone have any tips or techniques for doing differentiation under the integral sign to solve integrals? I'm just not sure how to look at them to figure out what to put in as a parameter.
Something like \[\frac{ d }{ dx } \int\limits_{a}^{x} f(t) dt\] ?
It depends on the integral. At this level of weird techniques, at least for me, there's not a single, easy to remember approach to take. Without a specific example, my general thought process is to see how a difficult integral differs from a simple one, and then try to insert a parameter to bridge the gap. Is there an example you have in mind?
Not exactly, I just heard about it from a book I was reading and I thought it would be fun to learn another method of solving integrals. I guess I might just have to kind of figure it out by just trying to find problems and solving them. Do you know any places I might be able to find some integrals to practice on?
if \[F(x)=\int\limits_{a}^{g(x)}f(t)dt\] then\[\large F'(x)=f(g(x))*g'(x)\] more generally if \[\Large F(x)=\int\limits_{g_{1}(x)}^{g_{2}(x)}f(t)dt\] \[\Large F'(x)=f(g_{2}(x))*g'_{2}(x)-f(g_{1}(x))*g'_{1}(x)\]
Not what I'm looking for, look here: http://fy.chalmers.se/~tfkhj/FeynmanIntegration.pdf and http://planetmath.org/DifferentiationUnderIntegralSign.html are what I'm talking about
http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf
Yeah I read that earlier too. But I'm afraid I don't really know how to "see" it like I might an integration by parts or u-substitution. Like, I'll look for something with its derivative when doing u-sub, but what exactly am I looking for to do feynman integration?
I use them all the time with Gaussian integrals: \[ \int dx \space e^{-x^2} = \sqrt{\pi}\] What about \[ \int dx \space x^2 e^{-x^2} \]? introduce a parameter, lambda: \[\int dx \space x^2 e^{-\lambda x^2} = \int dx \frac{-\partial}{\partial \lambda}e^{-\lambda x^2} = -\frac{d}{d\lambda} \int dx \space e^{-\lambda x^2} = \frac{-d}{d\lambda} \sqrt{\pi / \lambda}\] Then set lambda equal to 1 after everything is finished. If you understand the technique it's faster than integration by parts, especially since you'd have to do it twice to arrive at that integral (Not actually true, there's a clever way to do this particular integral using only a single integration by parts but that's beside the point, because for x^4 and x^6 it's obviously much faster). The same idea can be used anytime you have powers of x that appear outside of functions like sin or exp.
Now I like this but I'm not sure I understand how it works. I'm unfamiliar with this notation of putting the stuff being integrated to the right of "dx" even though it seems to be exactly the same.
It doesn't matter. It's purely notational and doesn't mean anything special, put it back on the other side if you want.
It's a notational habit born of treating \[\int dx \] as a self-contained operator.
Gotcha, alright cool I'm perfectly fine with that. But where does the constant of integration disappear to?
there isn't one, it's a definite integral. I didn't write the limits because I assumed you were familiar with the Gaussian integral: \[ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi} \]
there should be a constant that you need to find it.. http://www.youtube.com/watch?v=EgitbLcCG-I
What are you talking about? That is a perfectly well defined integral without any constants needed...
Oh. I see. The approach in that video is not the same as the approach I described.
I need to learn how to do integration as well as you guys.
It just takes practice. It only seems easy to some of us because we live and breathe calculus every day of our lives :)
@Jemurray3 What are hints that you'd want to use this method?
Basically the hints that you want to use this method is when every other method you try doesn't work lol.
Most commonly in my experience with it you may find it to be a much faster way to arrive at answers rather than repeated integration by parts. It's a good trick to know, in general. There related idea of differentiation under the integral sign is useful also.