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Let $ \gamma (t)= e^{it} $ where $0 \leq t \leq 2 \pi.$

Evaluate $\int_{\gamma}$ $e^{z}$ $dz$ .

Use the result to show that $\int_{0}^{2\pi} e^{\cos(t)}\cos(t+ \sin(t)) dt = 0$.

I have worked out the first contour integral to be 0 but I am unsure of how the result can help in showing the other integral.

blue
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  • If you want to see it evaluated without complex analysis, check out this post which ironically wasn't posted that long ago. – dustin Nov 20 '14 at 21:36

1 Answers1

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Write down direct value of integral using parametrization:

$$\oint_{\gamma}e^{z}dz=\int_{0}^{2\pi}e^{e^{it}}ie^{it}dt=i\int_{0}^{2\pi}e^{e^{it}+it}dt=\\=i\int_{0}^{2\pi}e^{\cos t+i\sin(t)+it}dt$$

But:

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

So:

$$i\int_{0}^{2\pi}e^{\cos t+i\sin(t)+it}dt=i\int_{0}^{2\pi}e^{\cos(t)}\cos(t+\sin(t))dt-\int_{0}^{2\pi}e^{\cos(t)}\sin(t+\sin(t))dt$$

So $0=\Im(\oint_{\gamma}e^{z}dz)=\int_{0}^{2\pi}e^{\cos(t)}\cos(t+\sin(t))dt$.

agha
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  • this helps a lot thank you but could I just ask where the second integral goes? the one with the sin(t+sin(t))? – blue Nov 20 '14 at 21:54
  • Yes, simply put $e^{i(t+\sin t)}=\cos(t+\sin(t))+i\sin(t+\sin(t))$ and use fact, that $i^2=-1$. – agha Nov 20 '14 at 21:56