the integral a complex variable

Question

$\int\limits_{0}^{2\pi}e^{\cos\varphi}(\cos\varphi-\sin\varphi)d\varphi$

I think

$e^{i\varphi}=z$ $\to d\varphi=\frac{dz}{iz}$

$\cos\varphi=\frac{z^2+1}{2z}$

$\sin\varphi=\frac{z^2-1}{2iz}$

$\oint\limits_{|z|=1}^{}e^{\frac{z^2+1}{2z}}\left(\frac{z^2+1}{2z}-\frac{z^2-1}{2iz}\right)dz$

$z=0$ - essential singularity

We need to be expanded in a Laurent but I do not know how there are so many polchaetsya values at $\frac{1}{z}$

Please don't number questions i.e. "complex integration 5". They do not make good titles. — , Dec 10 '15 at 05:45
The integral $\int_0^{2\pi} e^{\cos \phi},\cos \phi=2\pi I_1(1)$, where $I_1(z)$ is the First Kind Modified Bessel Function of order $1$. — Mark Viola, Dec 10 '15 at 05:55

Yuri Negometyanov · Answer 1 · 2015-12-10T10:43:03.360

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Mark that $$\int_0^{2\pi}e^{cos \phi}\sin\phi d\phi =\int_1^1e^ydy=0 \text{ (no singularities)},$$ so the artifical transformations $$\int_0^{2\pi}e^{cos \phi}(\cos\phi-\sin\phi) d\phi = \int_0^{2\pi}e^{cos \phi}(\cos\phi + i\sin\phi) d\phi = \int_0^{2\pi}e^{cos \phi}e^{i\phi} d\phi$$ easy the task

edited Dec 10 '15 at 10:43

answered Dec 10 '15 at 10:37

Yuri Negometyanov

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not quite sure where it came from $+isin\varphi$ – Olya12 Dec 10 '15 at 12:57
we can add zero – Yuri Negometyanov Dec 10 '15 at 18:39

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

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$$\int_0^{2\pi}e^{\cos\varphi}(\cos\varphi-\sin\varphi)~d\varphi~=~2\int_0^\pi e^{\cos\varphi}\cos\varphi~d\varphi~=~2\pi~I_1(1).$$

See Bessel function for more information.

edited Apr 13 '17 at 12:20

Community

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answered Dec 10 '15 at 11:58

Lucian

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did not quite understand what $I_1(1)$ – Olya12 Dec 10 '15 at 12:57
@Olya12: Click the links and find out... – Lucian Dec 10 '15 at 13:14

the integral a complex variable

2 Answers2