Contour Integrals (to Evaluate Real Integrals)

**roninpro** · January 17th 2011, 06:28 PM

Hello. I am attempting to evaluate

$\displaystyle \int_0^\infty \frac{x^{1/4}}{1+x^3}$

using a contour integral on the function $\displaystyle \frac{e^{(1/4)\log z}}{1+z^3}$ . My contour will be a "keyhole contour" suggested by Wikipedia: Methods of contour integration - Wikipedia, the free encyclopedia.

I tried to follow their example, but could not understand this:

$\displaystyle \int_R^\varepsilon {\sqrt{z} \over z^2+6z+8}\,dz=\int_R^\varepsilon {e^{{1\over 2} \mathrm{Log}(z)} \over z^2+6z+8}\,dz=\int_R^\varepsilon {e^{{1\over 2}(\log{|z|}+i \arg{z})} \over z^2+6z+8}\,dz=\int_R^\varepsilon { e^{{1\over 2}\log{|z|}}e^{1/2(2\pi i)} \over z^2+6z+8}\,dz$

Why is it permissible to substitute $\text{arg }z=2\pi$ ? Is it perhaps because they have taken $R\to \infty$ , so this line gets close to the $2\pi$ position?

I would appreciate any clarification on this example.

**Random Variable** · January 17th 2011, 09:07 PM

It's true when you let $\epsilon$ go to 0.

**roninpro** · January 17th 2011, 09:23 PM

Ah - I was personally using slightly different notation than Wikipedia, letting the little circle be described by $(1/R)e^{it}$ . In any case, I see what you mean.

Thanks for the reply.

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