How does WolframAlpha use the gamma function to solve this integral?

[Gavroy]

Hi

I have this integral that I want to express in terms of a gamma function. Unfortunately I am unable to bring it in this form. So can you give me a hint how wolframalpha does thishttp://m.wolframalpha.com/input/?i=∫e^(ix)/(ix)^(1/5)dx+from+-+infinity+to+infinity&x=10&y=3

 

[bossman27]

I don't see how it produced \Gamma (\frac{4}{5}), but I did find one way to solve it:

The Fourier transform of f(x) in non-unitary, angular frequency form is:

\hat f (\upsilon) = \int_{-\infty}^{\infty} f(x) e^{-i \upsilon x} dx

We have f(x) = (ix)^{-\frac{1}{5}} and will just need to evaluate \hat f (\upsilon) at \upsilon = -1 once we find an expression for it:

You'll note that the bottom entry in this table (http://en.wikipedia.org/wiki/Fourier_transform#Distributions) works for us in this situation, with \alpha = \frac{1}{5}

At \upsilon = -1 the expression reduces to \hat f (-1) = \frac{2 \pi}{\Gamma (\frac{1}{5})} = 1.368... -- agreeing with wolframalpha's calculated value.