zakinutuzi

2021-12-30

Solve integrals using residue theorem?

$\int}_{0}^{\pi}\frac{d\theta}{2+\mathrm{cos}\theta$

${\int}_{0}^{\mathrm{\infty}}\frac{x}{{(1+x)}^{6}}dx$

kalfswors0m

Beginner2021-12-31Added 24 answers

Hint for the First one First compute

$I={\int}_{0}^{2\pi}\frac{d\theta}{2+\mathrm{cos}\theta}={\int}_{0}^{2\pi}R(\mathrm{cos}\left(\theta \right),\text{}\mathrm{sin}\left(\theta \right))d\theta$

Where R is the rational function given by

$R(x,y)=\frac{1}{2+x}$

How to do this using the residue theorem? Put$z={e}^{it}=\mathrm{cos}\left(t\right)+i\mathrm{sin}\left(t\right)$ , thus

$\mathrm{cos}\left(t\right)=Re\left(z\right)=\frac{z+{z}^{-1}}{2}$

$\mathrm{sin}\left(t\right)=Im\left(z\right)=\frac{z-{z}^{-1}}{2i}$

$dz=i{e}^{it}dt=izdt\Rightarrow dt=\frac{1}{iz}dz$

Then I can be seen as a contour integral, solve it by using residues

$I={\int}_{0}^{2\pi}R(\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right))dt=$

${\int}_{\left|z\right|=1}R(\frac{z+\frac{1}{z}}{2},\frac{z-\frac{1}{z}}{2i})\frac{1}{iz}dz$

Hence in your case the integral you will compute is

$I={\int}_{\left|z\right|=1}\left(\frac{1}{2+\frac{z+\frac{1}{z}}{2}}\right)\frac{1}{iz}dz=\frac{2}{i}$

$\int}_{\left|z\right|=1}\frac{dz}{{z}^{2}+4z+1$

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

$\int}_{0}^{\pi}\frac{d\theta}{2+\mathrm{cos}\theta}=\frac{1}{i}{\int}_{\left|z\right|=1}\frac{dz}{{z}^{2}+4z+1$

Where R is the rational function given by

How to do this using the residue theorem? Put

Then I can be seen as a contour integral, solve it by using residues

Hence in your case the integral you will compute is

which can be easily obtain by the residue theorem!

Finally: Note that

and hence your result follows by computing the next integral

Linda Birchfield

Beginner2022-01-01Added 39 answers

For the second integral:

Note first that this integral is easily done by recognizing that$x=(1+x)-1$ , so the integral is really

$\int}_{0}^{\mathrm{\infty}}\frac{dx}{{(1+x)}^{5}}-{\int}_{0}^{\mathrm{\infty}}\frac{dx}{{(1+x)}^{6}}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20$

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

$\oint}_{C}dz\frac{z\mathrm{log}z}{{(1+z)}^{6}$

where C is a keyhole contour of outer radius R about the positive real axis. The contour integral is then equal to

$\int}_{\u03f5}^{R}dx\frac{x\mathrm{log}x}{{(1+x)}^{6}}+iR{\int}_{0}^{2\pi}d\theta {e}^{i\theta}\frac{R{e}^{i\theta}\mathrm{log}\left(R{e}^{i\theta}\right\}\left\{{(1+R{e}^{i\theta})}^{6}\right\}}{$

$+{\int}_{R}^{\u03f5}dx\frac{x(\mathrm{log}x+i2\pi )}{{(1+x)}^{6}}+i\u03f5{\int}_{2\pi}^{0}d\varphi$

$e}^{i\varphi}\frac{\u03f5{e}^{i\varphi}\mathrm{log}\left(\u03f5{e}^{i\varphi}\right)}{{(1+\u03f5{e}^{i\varphi})}^{6}$

As$R\to \mathrm{\infty}$ , the second integral vanishes as $\frac{\mathrm{log}R}{{R}^{4}}$ . As $\u03f5\to 0$ , the fourth integral vanishes as ${\u03f5}^{2}\mathrm{log}\u03f5$ . Thus, the contour integral is, in this limit

$-i2\pi {\int}_{0}^{\mathrm{\infty}}dx\frac{x}{{(1+x)}^{6}}$

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

$\int}_{0}^{\mathrm{\infty}}dx\frac{x}{{(1+x)}^{6}}=\frac{1}{20$

which agrees with the above.

Note first that this integral is easily done by recognizing that

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

where C is a keyhole contour of outer radius R about the positive real axis. The contour integral is then equal to

As

By the residue theorem, the contour integral is also equal to

Putting this altogether, we get that

which agrees with the above.

karton

Expert2022-01-09Added 613 answers

Hint for the first one:

Consider the function

and find its poles. Then use the known formula for residues:

Under the assumption that f has a pole of order m at

And finally, apply the Residue Theorem.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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