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Thread: Techniques of integration (1)

  1. #1
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    Techniques of integration (1)

    Evaluate $\displaystyle I(a)=\int_0^{\infty} e^{-a(x^2 + x^{-2})} \ dx, \ \ 0 < a \in \mathbb{R}.$
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  2. #2
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    Krizalid's Avatar
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    i'm quite rusty, it's been a long time i don't do many integration problems.

    it's $\displaystyle \int_{0}^{\infty }{e^{-a\left( x^{2}+x^{-2} \right)}\,dx}=\int_{0}^{1}{e^{-a\left( x^{2}+x^{-2} \right)}}+\int_{1}^{\infty }{e^{-a\left( x^{2}+x^{-2} \right)}},$ thus $\displaystyle \int_{1}^{\infty }{\left( 1+\frac{1}{x^{2}} \right)e^{-a\left( x^{2}+x^{-2} \right)}\,dx}=\int_{0}^{\infty }{e^{-a\left( x^{2}+2 \right)}\,dx},$ and your integral equals $\displaystyle \frac{1}{2e^{2a}}\sqrt{\frac{\pi }{a}}.$
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  3. #3
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    Quote Originally Posted by Krizalid View Post

    thus $\displaystyle \int_{1}^{\infty }{\left( 1+\frac{1}{x^{2}} \right)e^{-a\left( x^{2}+x^{-2} \right)}\,dx}=\int_{0}^{\infty }{e^{-a\left( x^{2}+2 \right)}\,dx},$
    good work! for those who don't see where the above equality came from: in the left hand side substitute $\displaystyle x - x^{-1}=t.$
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