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I want to evaluate the integral $$\int_{\gamma} \sin{(2z)} \ {\rm d}z$$ where $\gamma$ is the line segment joining the point $i+1$ to the point $-i$.

Thus $\gamma(t) = -i+t(2i+1)$ for $0\le t\le1$.

So I want to calculate \begin{align}\int_{\gamma} \sin{(2z)} \ {\rm d}z &=\int^{1}_{0} f(\gamma(t))\gamma'(t) \ {\rm d}t\\ &=\int_{0}^{1} \sin{[2(-i+t(1+2i))]}(1+2i) \ {\rm d}t \\ &=(1+2i)\int^{1}_{0} \sin{[2t+i(4t-2)]} \ {\rm d}t \\ &= (1+2i)\int^{1}_{0} \sin{(2t)}\cosh{(2-4t)}-i\cos{(2t)}\sinh{(2-4t)} \ {\rm d}t \\ &=(1+2i)\left[\int^{1}_{0}\sin{(2t)}\cosh{(2-4t)} \ {\rm d}t\, - i\int^{1}_{0}\cos{(2t)\sinh{(2-4t) \ {\rm d}t}}\right]\end{align}

Now this seems extremely long winded, is there any other way to calculate this?

dustin
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user2850514
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    In fact there is. Use this. – Git Gud Jan 01 '15 at 23:59
  • Did you get $\dfrac{1}{2}(\cos(2+2i)-\cos(-2i))$ as the answer? If so, that suggests a much simpler method. – JimmyK4542 Jan 01 '15 at 23:59
  • @GitGud of course I shall use the fundamental theorem of calculus applied to contour integrals. – user2850514 Jan 02 '15 at 00:09
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    Also, if you want to chug through the calculation, I'd recommend partitioning your $\sin$ function into $\sin[(2+4i)t+(-2i)]=\sin[(2+4i)t]\cos(-2i)+\cos[(2+4i)t]\sin(-2i)$ which would be a bit easier to directly compute than what you have. – Darrin Jan 02 '15 at 00:09

1 Answers1

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We have $\gamma(t)=-i+t(2i+1)$ for $0\le t\le 1$. Since $\gamma$ is smooth and $f(z) = \sin{(2z)}$ is continuous, let $F = \int f$ and note $\gamma(1)=1+i$, $\gamma(0)=-i$. By the fundamental theorem of calculus applied to contour integrals

$$\int_{\gamma} f = F(\gamma(1))-F(\gamma(0)).$$

Therefore \begin{align}\int_{\gamma} \sin{(2z)} \ {\rm d}z &= -\frac{1}{2}\cos{(2(1+i))}+\frac{1}{2}\cos{(2(-i))} \\ &= \frac{1}{2}\left[\cos{(2i)-\cos{(2+2i)}}\right].\end{align}

user2850514
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