Techniques of integration (1)

**NonCommAlg** · April 24th 2009, 05:50 PM

Evaluate $I(a)=\int_0^{\infty} e^{-a(x^2 + x^{-2})} \ dx, \ \ 0 < a \in \mathbb{R}.$

**Krizalid** · April 24th 2009, 06:35 PM

i'm quite rusty, it's been a long time i don't do many integration problems.

it's $\int_{0}^{\infty }{e^{-a\left( x^{2}+x^{-2} \right)}\,dx}=\int_{0}^{1}{e^{-a\left( x^{2}+x^{-2} \right)}}+\int_{1}^{\infty }{e^{-a\left( x^{2}+x^{-2} \right)}},$ thus $\int_{1}^{\infty }{\left( 1+\frac{1}{x^{2}} \right)e^{-a\left( x^{2}+x^{-2} \right)}\,dx}=\int_{0}^{\infty }{e^{-a\left( x^{2}+2 \right)}\,dx},$ and your integral equals $\frac{1}{2e^{2a}}\sqrt{\frac{\pi }{a}}.$

**NonCommAlg** · April 24th 2009, 06:50 PM

Originally Posted by Krizalid

thus $\int_{1}^{\infty }{\left( 1+\frac{1}{x^{2}} \right)e^{-a\left( x^{2}+x^{-2} \right)}\,dx}=\int_{0}^{\infty }{e^{-a\left( x^{2}+2 \right)}\,dx},$

good work!

for those who don't see where the above equality came from: in the left hand side substitute $x - x^{-1}=t.$

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