# Solve integrals using residue theorem? \int_0^{\pi}\frac{d\theta}{2+\cos\theta} \int_0^\infty\frac{x}{(1+x)^6}dx

Solve integrals using residue theorem?
${\int }_{0}^{\pi }\frac{d\theta }{2+\mathrm{cos}\theta }$
${\int }_{0}^{\mathrm{\infty }}\frac{x}{{\left(1+x\right)}^{6}}dx$
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kalfswors0m
Hint for the First one First compute

Where R is the rational function given by
$R\left(x,y\right)=\frac{1}{2+x}$
How to do this using the residue theorem? Put $z={e}^{it}=\mathrm{cos}\left(t\right)+i\mathrm{sin}\left(t\right)$, thus
$\mathrm{cos}\left(t\right)=Re\left(z\right)=\frac{z+{z}^{-1}}{2}$
$\mathrm{sin}\left(t\right)=Im\left(z\right)=\frac{z-{z}^{-1}}{2i}$
$dz=i{e}^{it}dt=izdt⇒dt=\frac{1}{iz}dz$
Then I can be seen as a contour integral, solve it by using residues
$I={\int }_{0}^{2\pi }R\left(\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right)\right)dt=$
${\int }_{|z|=1}R\left(\frac{z+\frac{1}{z}}{2},\frac{z-\frac{1}{z}}{2i}\right)\frac{1}{iz}dz$
Hence in your case the integral you will compute is
$I={\int }_{|z|=1}\left(\frac{1}{2+\frac{z+\frac{1}{z}}{2}}\right)\frac{1}{iz}dz=\frac{2}{i}$
${\int }_{|z|=1}\frac{dz}{{z}^{2}+4z+1}$
which can be easily obtain by the residue theorem!
Finally: Note that
${\int }_{0}^{\pi }\frac{d\theta }{2+\mathrm{cos}\theta }=\frac{I}{2}$
and hence your result follows by computing the next integral
${\int }_{0}^{\pi }\frac{d\theta }{2+\mathrm{cos}\theta }=\frac{1}{i}{\int }_{|z|=1}\frac{dz}{{z}^{2}+4z+1}$
###### Not exactly what you’re looking for?
Linda Birchfield
For the second integral:
Note first that this integral is easily done by recognizing that $x=\left(1+x\right)-1$, so the integral is really
${\int }_{0}^{\mathrm{\infty }}\frac{dx}{{\left(1+x\right)}^{5}}-{\int }_{0}^{\mathrm{\infty }}\frac{dx}{{\left(1+x\right)}^{6}}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}$
One may also use the residue theorem. However, one must choose an appropriate contour and integrand. In this case, a useful contour integral to consider is
${\oint }_{C}dz\frac{z\mathrm{log}z}{{\left(1+z\right)}^{6}}$
where C is a keyhole contour of outer radius R about the positive real axis. The contour integral is then equal to
${\int }_{ϵ}^{R}dx\frac{x\mathrm{log}x}{{\left(1+x\right)}^{6}}+iR{\int }_{0}^{2\pi }d\theta {e}^{i\theta }\frac{R{e}^{i\theta }\mathrm{log}\left(R{e}^{i\theta }\right\}\left\{{\left(1+R{e}^{i\theta }\right)}^{6}\right\}}{}$
$+{\int }_{R}^{ϵ}dx\frac{x\left(\mathrm{log}x+i2\pi \right)}{{\left(1+x\right)}^{6}}+iϵ{\int }_{2\pi }^{0}d\varphi$
${e}^{i\varphi }\frac{ϵ{e}^{i\varphi }\mathrm{log}\left(ϵ{e}^{i\varphi }\right)}{{\left(1+ϵ{e}^{i\varphi }\right)}^{6}}$
As $R\to \mathrm{\infty }$, the second integral vanishes as $\frac{\mathrm{log}R}{{R}^{4}}$. As $ϵ\to 0$, the fourth integral vanishes as ${ϵ}^{2}\mathrm{log}ϵ$. Thus, the contour integral is, in this limit
$-i2\pi {\int }_{0}^{\mathrm{\infty }}dx\frac{x}{{\left(1+x\right)}^{6}}$
By the residue theorem, the contour integral is also equal to $i2\pi$ times the residue at the pole $x={e}^{i\pi }$. (Note how important it is to get the argument correct.) The residue at this pole is
$\frac{1}{5!}{\left[\frac{{d}^{5}}{{dz}^{5}}\left(z\mathrm{log}z\right)\right]}_{z={e}^{i\pi }}=-\frac{3!}{5!}$
Putting this altogether, we get that
${\int }_{0}^{\mathrm{\infty }}dx\frac{x}{{\left(1+x\right)}^{6}}=\frac{1}{20}$
which agrees with the above.
###### Not exactly what you’re looking for?
karton

Hint for the first one:
Consider the function
$f\left(z\right)=\frac{2}{{z}^{2}+4z+1}$
and find its poles. Then use the known formula for residues:
Under the assumption that f has a pole of order m at ${z}_{0}$
$Res\left({z}_{0},f\right)=\frac{1}{\left(m-1\right)!}\underset{z\to {z}_{0}}{lim}\frac{{d}^{m-1}}{d{z}^{m-1}}\left(\left(z-{z}_{0}{\right)}^{m}f\left(z\right)\right).$
And finally, apply the Residue Theorem.