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Math Help - Integration

  1. #1
    Junior Member
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    Integration

    Could you help me with this?

    By considering the integral of  e^{-yx^2} over a suitable region of  \mathbb{R}^2 , show that for  0<a<b<\infty ,

     \int_{0}^{\infty} \frac{e^{-ax^2} - e^{-bx^2}}{x^2} dx = \sqrt{\pi} ( \sqrt{b} - \sqrt {a} )
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    the hint was already given, we have that \frac{e^{-ax^2}-e^{-bx^2}}{x^2}=\int_a^b e^{-tx^2}\,dt, now reverse integration order (justified by Tonelli's Theorem since the integrand is non negative, becuase of b>a) and use the fact that \int_{x\ge0}e^{-x^2}\,dx=\frac{\sqrt\pi}2, hence, the result.
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