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In Titchmarsh's book on Fourier Transforms it states that the integral

$$\int\nolimits_0^\infty \frac{x^{-a}}{1+x^2} dx,\quad 0<a<1$$

may be calculated by either contour integration or by series expansion. It certainly is easy to do this, technically. However, since the series for $\frac{1}{1+x^2}$ will only have radius of convergence equal to 1, due to the poles at $\pm i$, how does one justify doing the expansion then the integration?

I know this must be simple but I'm not sure why it is true. Is it because the terms when combied with $x^{-a}$ form an absolutely convergent series?

Thanks, Tom

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Perhaps Titchmarsh meant this:

$$\int^{\infty}_0 \frac{ x^{-a} }{1+x^2} dx = \int^{1}_0 \frac{ x^{-a} }{1+x^2} dx +\int^{\infty}_1 \frac{ x^{-a} }{1+x^2} $$

Let $x=\frac{1}{u}$ into the second integral and we arrive at: $$\int^{\infty}_0 \frac{ x^{-a} }{1+x^2} = \int^1_0 \frac{ x^a+x^{-a} }{1+x^2} dx $$

Now the series expansion of $\frac{1}{1+x^2}$ is valid over the domain of integration.

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Thanks - I've seen this method before and had forgotten it! – Tom Dickens Aug 30 '11 at 0:03

Use $\frac{1}{1+x^2} = 1 - x^2 + x^4 - \ldots$ on $(0,1)$ and $\frac{1}{1+x^2} = \frac{1}{x^2} \frac{1}{1+x^{-2}} = \frac{1}{x^2} - \frac{1}{x^4} + \frac{1}{x^6} - \ldots$ on $(1,\infty)$.

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Thanks, this is a very useful kind of transformation to keep in mind! – Tom Dickens Aug 30 '11 at 0:06

You can use this path enter image description here

The singular points are $e^{i\left( {\pi + 2k\pi } \right)/s} $ and only $e^{i\pi /s} $ is inside the path By residue theorem: $$\int\limits_0^R {\frac{1}{{1 + x^s }}dx + \int\limits_{\Gamma _R }^{} {\frac{1}{{1 + z^s }}dz - e^{2i\pi /s} } } \int\limits_0^R {\frac{1}{{1 + x^s }}dx = 2\pi i\frac{1}{{se^{i\left( {s - 1} \right)\pi /s} }}} $$ The path in the circunference tend to 0 ad R tend to infinitys,so: $$\int\limits_0^\infty {\frac{1}{{1 + x^s }}dx = - \frac{{2\pi i}}{s}\frac{{e^{i\pi /s} }}{{1 - e^{2i\pi /s} }}} = \frac{\pi }{s}\frac{1}{{\sin \left( {\frac{\pi }{s}} \right)}}$$ You can also calculate this integral from the previous integral: $$\int\limits_0^\infty {\frac{{x^\alpha }}{{1 + x^\beta }}} dx = \frac{1}{{\alpha + 1}}\int\limits_0^\infty {\frac{{\left( {1 + \alpha } \right)x^\alpha }}{{1 + \left( {x^{1 + \alpha } } \right)^{\frac{\beta }{{1 + \alpha }}} }}} dx = \frac{\pi }{\beta }\frac{1}{{\sin \left( {\frac{{\left( {1 + \alpha } \right)\pi }}{\beta }} \right)}}$$

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Sorry but i don't know how to calculate this integral by series. I hope this will be useful for you ,despite the fact it doesn't use series – Pablo Sep 12 '11 at 21:21

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