By using Fourier transform methods compute the integral $$\int_{-\infty}^\infty \frac{1}{x^4 + 7x^2 + 6} \ dx.$$
I know that I should use the inversion formula and evaluate it in some point, but I can't figure out how to do this.
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If that helps the integral equals $\frac{\pi}{5} (1 - \frac{1}{\sqrt{6}})$ but I calculated it using residues. – Giorgos Giapitzakis Jun 22 '21 at 13:23
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You can also calculate the integral using partial fractions (just as an alternative). – Clayton Jun 22 '21 at 13:43
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3Note that $$\frac{1}{x^4+7x^2+6}=\frac{1}{(x^2+1)(x^2+6)},$$ and see Compute $\int_{0}^{\infty} \frac{dx}{(a^2+x^2)(b^2+x^2)} $ using fourier transform of $e^{-a|x|}$ or How do I show $\int_{-\infty}^\infty \frac 1{(a^2+s^2)(b^2+s^2)} ds=\frac {\pi}{ab(a+b)}$ using the solution to the following Fourier transform?. – projectilemotion Jun 22 '21 at 13:58
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Okay, this was absolutely useful. Thank you. – Gabrielek Jun 22 '21 at 14:26