Abstract Nonsense

Crushing one theorem at a time

Another integral


Here is a semi-interesting integral

Problem: Compute \displaystyle J=\int_0^{\infty}e^{-(x^2+\tfrac{1}{x^2})}dx

Solution:Clearly

\displaystyle J=\int_0^1 e^{-(x^2+\tfrac{1}{x^2})}dx+\int_1^{\infty}e^{-(x^2+\tfrac{1}{x^2})}dx

Let x=\frac{1}{z}\implies dx=\frac{-dz}{z^2} in the first to see that

\displaystyle J=\int_0^{\infty}\frac{1}{x^2}e^{-(x^2+\tfrac{1}{x^2})}+\int_1^{\infty}e^{-(x^2+\tfrac{1}{x^2})}dx

Which upon combination gives

\displaystyle J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+\tfrac{1}{x^2})}dx

Note though that

\displaystyle x^2+\frac{1}{x^2}=x^2-2x\frac{1}{x}+\frac{1}{x^2}-2=\left(x-\frac{1}{x}\right)^2-2

So, letting \displaystyle z=x-\frac{1}{x}\implies dz=1+\frac{1}{x^2}dx we see that

\displaystyle J=\int_0^{\infty}e^{-z^2-2}dz=\frac{\sqrt{\pi}}{2e^2}

\blacksquare

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May 29, 2010 - Posted by | Uncategorized

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