Assuming that the integrals below exist, $$\int_{-\infty}^{\infty} f(x-\frac{1}{x})dx=\int_{-\infty}^{\infty}f(x)dx$$ Use the above relation to evaluate the following integral: $$\int_{-\infty}^{\infty} e^{-(x^2-2)^2-(\frac{1}{x^2}-2)^2}dx$$ Express the result in terms of the gamma function.\
I have tried expanding the terms in the exponent and factoring out a constant but i can't seem to get it in terms of a gamma function. Here's what i've tried so far:\
Exp$[-(x^2-2)^2-(\frac{1}{x^2}-2)^2]$=Exp$[-x^4+4x^2-\frac{1}{x^4}+\frac{4}{x^2}-8]$\
Now rearranging terms and factoring out the constant gives: $$e^{-8}\int_{\infty}^{\infty}e^{-(x^4+\frac{1}{x^4})}e^{4(x^2+\frac{1}{x^2})}dx$$ Next i said: $(x^4+\frac{1}{x^4})=(x^2+\frac{1}{x^2})^2-2$. So we can rewrite out integral as:
$$e^{-8}e^{-2}\int_{-\infty}^{\infty}e^{-(x^2+\frac{1}{x^2})^2}e^{4(x^2+\frac{1}{x^2})}dx=e^{-10}\int_{-\infty}^{\infty}e^{-(x^2+\frac{1}{x^2})^2+4(x^2+\frac{1}{x^2})}dx$$
I'm not sure where to go from here. Can i use that given relation and make substitution $u=x^2-\frac{1}{x^2}$? Then i would have:
$$e^{-10}\int_{-\infty}^{\infty}e^{-u^2-4u}du$$
This doesn't seem right... \
The other thing i tried is to let $u=x^2$ so that $x=u^{1/2}$ and $dx=\frac{1}{2}u^{1/2-1}$. In this case the integral can be written as:
$$\frac{e^{-10}}{2}\int_{-\infty}^{\infty}e^{-(u+\frac{1}{u})^2+4(u+\frac{1}{u})}u^{1/2-1}du$$
Now we have it sort of looking like the gamma function but i don't know what to do with all the terms over "e".
Let $g(x) = (x^2-2)^2+(x^{-2}-2)^2$. Observe that $$g(x) = (x - x^{-1})^4 + 2,$$ so that the choice $f(x) = e^{-(x^4+2)}$ gives $$\int_{x=-\infty}^\infty e^{-g(x)} \, dx = e^{-2} \int_{x=-\infty}^\infty \exp(-(x-x^{-1})^4) \, dx = e^{-2} \int_{x=-\infty}^\infty e^{-x^4} \, dx.$$ Then all that remains is to perform the final integration. To this end, consider the substitution $$u = x^m, \quad x = u^{1/m}, \quad dx = \frac{1}{m} u^{-1+1/m} \, du$$ for $m > 0$. Then $$\int_{x=0}^\infty e^{-x^m} \, dx = \frac{1}{m} \int_{u=0}^\infty u^{-1+1/m} e^{-u} \, du = \frac{1}{m} \Gamma(1/m) = \Gamma(1 + 1/m).$$ For $m = 4$, we finally obtain the desired result $$2 e^{-2} \Gamma(5/4).$$