Integral of fourier transform of $\frac{1}{1+x^8}$

Question

How to calculate $\int\mathcal{F}\left( \frac{1}{1+x^8} \right)$? Are there special hits using the fact that it's Fourier transform?

Answer 1

Defining the Fourier transform as $$ \mathcal{F}\{f(x)\}=\hat{f}(\xi) = \int_{-\infty}^\infty f(x)\ e^{- 2\pi i x \xi}\,dx, $$ and the inverse transform as $$ \mathcal{F}^{-1}\{\hat f(x)\}=f(x) = \int_{-\infty}^\infty \hat f(\xi)\ e^{2 \pi i \xi x}\,d\xi $$ you have that $$ f(0)=\int_{-\infty}^\infty \hat f(\xi)\,d\xi $$ and then $$ \int_{-\infty}^{\infty} \mathcal F\left\{\frac{1}{1+x^8}\right\}\,d\xi=\left.\frac{1}{1+x^8}\right|_{x=0}=1 $$ Using the definition $\hat{f}(\xi) = \int_{-\infty}^\infty f(x)\ e^{- i x \xi}\,dx$ or $\hat{f}(\xi) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)\ e^{- i x \xi}\,dx$, you'll find $2\pi$ or $\sqrt{2\pi}$.