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    I've searched for this on the Internet, and I'm struggling to understand it. Most explanations are referring to partial derivatives or multivariable functions.

    Does the method use the result \displaystyle \frac{\text{d}}{\text{d}x} \int_a^x \text{f}(t) \, \text{d}t = \text{f}(x)?

    So, for an example, would it be possible to find \displaystyle \int_0^{\infty} e^{-t^2} \, \text{d}t this way? (And if so, how?)
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    (Original post by Glutamic Acid)
    I've searched for this on the Internet, and I'm struggling to understand it. Most explanations are referring to partial derivatives or multivariable functions.

    Does the method use the result \displaystyle \frac{\text{d}}{\text{d}x} \int_a^x \text{f}(t) \, \text{d}t = \text{f}(x)?
    Not in general, although you might use that as well.

    The idea of differentiating under the integral sign is that under suitable circumstances,

    \frac{d}{da} \int_A^B f(x,a) \,dx = \int_A^B \frac{d}{da} f(x,a) \,dx

    e.g. \int_0^\infty e^{-ax} \, dx = 1/a.

    So the rule above gives \int_0^\infty \frac{d}{da} e^{-ax} \,dx = \frac{d}{da} 1/a

    That is, \int_0^\infty -x e^{-ax} \,dx = -1/a^2.

    Strictly speaking, the differentiation w.r.t. a is a partial derivative.

    So, for an example, would it be possible to find \displaystyle \int_0^{\infty} e^{-t^2} \, \text{d}t this way? (And if so, how?)
    It is possible, but it isn't easy, and it's very unobvious.

    See Q14 on http://www.damtp.cam.ac.uk/user/examples/A3a.pdf
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    (Original post by yusufu)
    What is your question?
    I don't quite follow; there are three.
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    (Original post by Glutamic Acid)
    I don't quite follow; there are three.
    My bad. I misunderstood your post.
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    (Original post by DFranklin)
    Not in general, although you might use that as well.

    The idea of differentiating under the integral sign is that under suitable circumstances,

    \frac{d}{da} \int_A^B f(x,a) \,dx = \int_A^B \frac{d}{da} f(x,a) \,dx

    e.g. \int_0^\infty e^{-ax} \, dx = 1/a.

    So the rule above gives \int_0^\infty \frac{d}{da} e^{-ax} \,dx = \frac{d}{da} 1/a

    That is, \int_0^\infty -x e^{-ax} \,dx = -1/a^2.

    Strictly speaking, the differentiation w.r.t. a is a partial derivative.
    Thanks, I think I get it, it was the step of putting d/da inside the integral which I was missing.

    So, just to check:
    \displaystyle \int_0^{\pi} \cos(ax) \, \text{d}x = \frac{\sin(a \pi)}{a}.

    \Rightarrow \displaystyle \int_0^{\pi} \frac{\text{d}}{\text{d}a} \cos(ax) \, \text{d}x = \frac{\pi \cos(a \pi)}{a} - \frac{\sin(a \pi)}{a^2}.

    \Rightarrow \displaystyle \int_0^{\pi} x \cos(ax) \, \text{d}x = \frac{\pi \cos(a \pi)}{a} - \frac{\sin(a \pi)}{a^2}.

    But this seems to break down even when a = 1. Or is this a case where differentiating under the integral won't work?
    It is possible, but it isn't easy, and it's very unobvious.

    See Q14 on http://www.damtp.cam.ac.uk/user/examples/A3a.pdf
    I see. There's no way I'd have thought of it, and I doubt I can prove it anyway.
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    d/da cos(ax) = -x sin(ax), surely?
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    Oh, haha, silly mistake. Thanks.
 
 
 
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