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Math Help - computing integral

  1. #1
    makoley4
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    computing integral

    i'm having a problem computing the integral below. could someone please assist me?

    integral of exp(-x^2 + xz -1/4z^2) dx
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  2. #2
    MHF Contributor
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    Small wonder you are having difficulty. That is non-trivial, to be sure. I have to ask a few things.

    1) Why do you think you need to compute this integral?

    2) Why do you think the answer exists in any convenient form?

    3) Do you get to use Complex Analysis? Actually, is this the original intent?

    4) Are you comfortable with Erf?

    5) If you are far enough along in mathematics to be contemplating such an integral, you should have learned to pay much better attention to your notation. It is not clear at all what your intent is in that exponent. I took it to mean a quadratic in x all divided by 4z^2. That is absolutely NOT what you have writtten. What did you really mean?
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by makoley4 View Post
    i'm having a problem computing the integral below. could someone please assist me?

    integral of exp(-x^2 + xz -1/4z^2) dx
    Of course, this makes sense if x\in(-\infty,\infty).

    - \left( {x^2 - xz + \frac{1}<br />
{4}z^2 } \right) = - \left( {x - \frac{1}<br />
{2}z} \right)^2 .

    So

    \int_{ - \infty }^{ + \infty } {e^{ - \left( {x - z/2} \right)^2 } \,dx} .

    Now make a substitution according to u=x-\frac z2. Sure, we'll get the beautiful result

    \int_{ - \infty }^{ + \infty } {e^{ - x^2 } \,dx} = 2\int_0^\infty {e^{ - x^2 } \,dx} = \sqrt \pi .

    The last result can be proven applyin' polar transform or nice play with double integration.
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  4. #4
    Forum Admin topsquark's Avatar
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    6) And do you have integration limits?

    -Dan
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TKHunny View Post
    4) Are you comfortable with Erf?
    what's that?
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    Look at for Error Function.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Look at for Error Function.
    oh! oh yeah, it's the error function. i've heard of that
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