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    Can you integrate x√lnx ? Or is this one of those that requires some other kind of method?

    Thanks
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    If you use u = ln x you get

    x\sqrt{\ln x}dx = \sqrt{u}e^{2u}du

    Then using v = \sqrt{u} you get

    2s^{2}e^{2s^{2}}ds

    This can't be done (leads to \int e^{x^{2}}dx integral, known to not be possible analytically with nice functions) so I think the original integral can't be done.
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    try the substitution of u = rootlnx, however i am not 100% sure.
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    Oh i see.

    Thanks for that Alpha
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    (Original post by AlphaNumeric)
    If you use u = ln x you get

    x\sqrt{\ln x}dx = \sqrt{u}e^{2u}du

    Then using v = \sqrt{u} you get

    2s^{2}e^{2s^{2}}ds

    This can't be done (leads to \int e^{x^{2}}dx integral, known to not be possible analytically with nice functions) so I think the original integral can't be done.
    I fully agree with that, I got it down to integrals with \int e^{x^{2}}dx involved too
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    Im not sure what you guys are doing. Ur letting u=ln x. Okay i get that. Then u subsitute this into the equation to form x√u. Use du/dx to get 1/x so du = 1/x dx. Put this back into the equation to form x^2√u.

    Now im totally lost
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    (Original post by AlphaNumeric)
    This can't be done (leads to \int e^{x^{2}}dx integral, known to not be possible analytically with nice functions) so I think the original integral can't be done.
    What do you mean by 'nice' functions? Does that imply that there are functions you can do it with?
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    Hang on...is this Edexcel C4 stuff?
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    (Original post by tom the mathematician)
    Im not sure what you guys are doing. Ur letting u=ln x. Okay i get that. Then u subsitute this into the equation to form x√u. Use du/dx to get 1/x so du = 1/x dx. Put this back into the equation to form x^2√u.

    Now im totally lost
    \int x\sqrt{\ln x}dx

    Let u = \ln x so x = e^{u} and dx = e^{u}du so

    \int x\sqrt{\ln x}dx = \int e^{u}\sqrt{u}e^{u}du = \int \sqrt{u}e^{2u}du

    Let u= v^2 so du = 2v dv so

    \int \sqrt{u}e^{2u}du = \int 2v^{2}e^{v^{2}}dv
    (Original post by BovineBeast)
    What do you mean by 'nice' functions? Does that imply that there are functions you can do it with?
    Polynomials, exponentials, trig functions, logs, that kind of thing.

    As an example, consider \int_{0}^{x} e^{-x^{2}}dx \equiv \Phi(x)

    There you've got a function \Phi(x) which is the function which solves \frac{d}{dx}f(x) = e^{-x^{2}} so you can 'integrate' e^{-x^{2}} but that's just because you've defined a function which is the answer to your problem but not actually gained anything by it.
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    Mathematica confirms it cannot be integrated nicely.
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    (Original post by AlphaNumeric)
    \int \sqrt{u}e^{2u}du = \int 2v^{2}e^{v^{2}}dv
    Polynomials, exponentials, trig functions, logs, that kind of thing.

    As an example, consider \int_{0}^{x} e^{-x^{2}}dx \equiv \Phi(x)

    There you've got a function \Phi(x) which is the function which solves \frac{d}{dx}f(x) = e^{-x^{2}} so you can 'integrate' e^{-x^{2}} but that's just because you've defined a function which is the answer to your problem but not actually gained anything by it.
    Okay, so here's a question - can you find the exact area under an e^(-x^2) curve? or do you have to just approximate it?
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    (Original post by tom the mathematician)
    Hang on...is this Edexcel C4 stuff?
    No, all you need to know for C4 is:
     \int f'(x)e^{f(x)} dx = e^{f(x)}
     \int \frac{f'(x)}{f(x)} dx = ln(f(x))
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    (Original post by BovineBeast)
    Okay, so here's a question - can you find the exact area under an e^(-x^2) curve? or do you have to just approximate it?
    If no analytic solution exists, then iterative methods must be used, so the "Normal distribution" values you get in Stats are approximations, though you can get as accurate as you need given enough computer power (thousands of decimal places easily with home computers).

    You can find an exact analytic solution if the limits for your I = \int e^{-x^{2}}dx problem are 2 of the following 3 values, 0, \infty or -\infty because then you can use polar coordinates to do nifty change of variables on I^{2} = \int e^{-x^{2}}dx \int e^{-y^{2}}dy and then taking the square root. It's a method you'd typically meet in the first year of uni.
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    (Original post by AlphaNumeric)
    If no analytic solution exists, then iterative methods must be used, so the "Normal distribution" values you get in Stats are approximations, though you can get as accurate as you need given enough computer power (thousands of decimal places easily with home computers).

    You can find an exact analytic solution if the limits for your I = \int e^{-x^{2}}dx problem are 2 of the following 3 values, 0, \infty or -\infty because then you can use polar coordinates to do nifty change of variables on I^{2} = \int e^{-x^{2}}dx \int e^{-y^{2}}dy and then taking the square root. It's a method you'd typically meet in the first year of uni.

    You do indeed, and a bloody lovely proof it is. Who'd have thought that it'd be the square root of pi if the limits were +-infinity?
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    (Original post by JohnSPals)
    You do indeed, and a bloody lovely proof it is. Who'd have thought that it'd be the square root of pi if the limits were +-infinity?
    Oh? Do tell. I'm interested.
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    (Original post by BovineBeast)
    Oh? Do tell. I'm interested.
    You need to understand double integrals and using a change of variables, so I doubt you'd understand it... Sorry!
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    I = \int_{0}^{\infty}e^{-x^{2}}dx

    I^2 = \int_{0}^{\infty}e^{-x^{2}}dx \; \int_{0}^{\infty}e^{-y^{2}}dy

    Lets start messing with variables to get

    I^2 = \int_{0}^{\infty} \int_{0}^{\infty}e^{-(y^{2}+x^{2})}dydx

    Polar coords says x^{2}+y^{2} = r^{2} and if you draw the Argand diagram you also get the limits for \theta being 0 and \frac{\pi}{2}

    Go from here and get your answer. I'm too drunk to bother...
 
 
 

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