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I have this integral but I don't how to treat it because has an imaginary.

$$ \int_0^{2\pi} e^{im\phi}~d\phi$$

Thanks for the help.

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    Have you tried the substitution $\phi=z/im$? – 2017-02-15

3 Answers 3

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If you set

$$z=e^{im\phi}\implies dz=imzd\phi\implies d\phi=-\frac imdz$$

and your integral becomes a "regular"complex one

$$-\frac im\oint_{|z|=1} z\,dz=0$$

which is also the result you get directly:

$$\int_0^{2\pi} e^{im\phi}d\phi=\left.\frac1{im}e^{im\phi}\right|_0^{2\pi}=0$$

Note: Assuming $\;0\neq m\in\Bbb Z\;$ .

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    You should exclude the case $m=0$. – 2017-02-16
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    @ZaidAlyafeai Of course, thanks. Edited. – 2017-02-16
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Hint

If you don't want to deal with imaginary substitution , use the Euler formula

$$e^{ix} = \cos(x)+i\sin(x)$$

Then you should be able to work with each integral separately.

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    (+1) This is exactly the way I would have proceeded given the OP's doubts about the complex nature of the integrand. – 2017-02-15
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The imaginary here doesn't pose a problem since the integration variable $\phi$ stays real.

You integrate normally for $m\neq 0$ $$\int_0^{2\pi}e^{im\phi}d\phi=\left[\frac{e^{im\phi}}{im}\right]_0^{2\pi}=\frac{e^{i2m\pi}-1}{im}=\frac{\sin(2m\pi)}{m}-i\,\frac{\cos(2m\pi)-1}{m}$$

And for $m=0$

$$\int_0^{2\pi}e^{im\phi}d\phi=\int_0^{2\pi}d\phi=2\pi$$

It would be a completely different thing to integrate on a curve $\mathcal C$ the quantity $\int_{\mathcal C}e^{imz}dz$.