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Math Help - how to calculate this integral!!!

  1. #1
    Junior Member toraj58's Avatar
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    how to calculate this integral!!!

    how to calulate this :

    <br />
\int_{-\infty}^{\infty}e^{-x^2}\ dx<br />

    it seems easy but i wonder how it is tricky to solve it.
    i planned several approaches but all of them failed. i think the answer should be someting similar to  \pi .
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by toraj58 View Post
    how to calulate this :

    <br />
\int_{-\infty}^{\infty}e^{-x^2}\ dx<br />

    it seems easy but i wonder how it is tricky to solve it.
    i planned several approaches but all of them failed. i think the answer should be someting similar to  \pi .
    see post #2 here. you should be able to finish up. it shows you how to find \int_0^\infty e^{-x^2}~dx
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  3. #3
    Junior Member toraj58's Avatar
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    Smile

    yeah;
    that was cool solution.
    it was realy nice for me to use double integral trick and then benefit from solving the problem in Polar Coordinates that was undeniably easy.
    so the answer is : \sqrt\pi
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by toraj58 View Post
    yeah;
    that was cool solution.
    it was realy nice for me to use double integral trick and then benefit from solving the problem in Polar Coordinates that was undeniably easy.
    so the answer is : \sqrt\pi
    yes
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by toraj58 View Post
    how to calulate this :

    <br />
\int_{-\infty}^{\infty}e^{-x^2}\ dx<br />

    it seems easy but i wonder how it is tricky to solve it.
    i planned several approaches but all of them failed. i think the answer should be someting similar to  \pi .
    Its also not hard to show that \Gamma\left(\tfrac{1}{2}\right)=\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}, where \Gamma(x)=\int_0^{\infty}e^{-t}t^{x-1}\,dt

    --Chris
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  6. #6
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    A Cute Angle Moo's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Its also not hard to show that \Gamma\left(\tfrac{1}{2}\right)=\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}, where \Gamma(x)=\int_0^{\infty}e^{-t}t^{x-1}\,dt

    --Chris
    And how do you know that \Gamma(\tfrac 12)=\sqrt{\pi} ?

    _____________________________________
    Here is a few other ways.

    1.
    Consider g(x)=\int_0^1 \frac{e^{-(t^2+1)x^2}}{t^2+1} ~dt

    \bullet Find g'(x)

    Let f(x)=\int_0^x e^{-u^2} ~du
    \bullet Show that g'(x)=-2f'(x) f(x)

    \bullet Integrate this latter formula.

    \bullet Find an expression for g(x) with respect to f(x).

    \bullet Show that \lim_{x \to \infty} g(x)=0

    \bullet Conclude : \lim_{x \to +\infty} f(x)=\sqrt{\frac{\pi}{4}}

    Since the integrand is even, \int_{-\infty}^\infty e^{-t^2} ~dt=2 \int_0^\infty e^{-t^2} ~dt

    ---------------------------------------
    2.
    \bullet Show that \forall x \in ]-1,+\infty[,~  \ln(1+x) \leq x

    \bullet Deduce that \left(1-\frac{t^2}{n}\right)^n \leq e^{-t^2} \leq \left(1+\frac{t^2}{n}\right)^n

    \bullet Integrate the previous inequality and make substitutions to get, \forall n \in \mathbb{N}^* :
    \sqrt{n} \int_0^{\pi/2} \sin^{2n+1}(u) ~du \leq \int_0^{\sqrt{n}} e^{-t^2} ~dt \leq \sqrt{n} \int_0^{\pi /2} \sin^{2n-2}(u) ~du

    \bullet Admit the Wallis integral : \int_0^{\pi/2} \sin^p(u) ~du \to \sqrt{\frac{\pi}{2p}} \quad \text{ when } p \to \infty
    And find the value of \int_0^\infty e^{-t^2} ~dt
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  7. #7
    Junior Member toraj58's Avatar
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    Smile

    thanks for alternative solutions.
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