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.

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- Kainui

- schrodinger

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- anonymous

Something like \[\frac{ d }{ dx } \int\limits_{a}^{x} f(t) dt\]
?

- anonymous

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?

- Kainui

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?

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## More answers

- anonymous

if \[F(x)=\int\limits_{a}^{g(x)}f(t)dt\]
then\[\large F'(x)=f(g(x))*g'(x)\]
more generally
if
\[\Large F(x)=\int\limits_{g_{1}(x)}^{g_{2}(x)}f(t)dt\]
\[\Large F'(x)=f(g_{2}(x))*g'_{2}(x)-f(g_{1}(x))*g'_{1}(x)\]

- Kainui

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

- anonymous

http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf

- Kainui

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?

- anonymous

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.

- Kainui

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.

- anonymous

It doesn't matter. It's purely notational and doesn't mean anything special, put it back on the other side if you want.

- anonymous

It's a notational habit born of treating
\[\int dx \]
as a self-contained operator.

- Kainui

Gotcha, alright cool I'm perfectly fine with that. But where does the constant of integration disappear to?

- anonymous

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:
\[ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi} \]

- anonymous

there should be a constant that you need to find it..
http://www.youtube.com/watch?v=EgitbLcCG-I

- anonymous

What are you talking about? That is a perfectly well defined integral without any constants needed...

- anonymous

Oh. I see. The approach in that video is not the same as the approach I described.

- anonymous

I need to learn how to do integration as well as you guys.

- anonymous

It just takes practice. It only seems easy to some of us because we live and breathe calculus every day of our lives :)

- anonymous

@Jemurray3
What are hints that you'd want to use this method?

- Kainui

Basically the hints that you want to use this method is when every other method you try doesn't work lol.

- anonymous

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.

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