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I'm trying to explicitly evaluate the following integral $$ \int_{-\pi}^{\pi} \frac{\sin\left(e^{ix}\right)}{e^{ix}} dx $$

I checked on WolframAlpha that the value of the integral is $2 \pi$. Using this, I attempted the following.


I analyze the conjugate of the integral and see that $$ \overline{\int_{-\pi}^{\pi} \frac{\sin\left(e^{ix}\right)}{e^{ix}} dx} = \int_{-\pi}^{\pi} \frac{\overline{\sin\left(e^{ix}\right)}}{\overline{e^{ix}}} dx = \int_{-\pi}^{\pi} \frac{\sin\left(e^{-ix}\right)}{e^{-ix}} dx \overset{\color{blue}{u = -x}}{=}\int_{-\pi}^{\pi} \frac{\sin\left(e^{iu}\right)}{e^{iu}} du $$ which confirms to us that the integral is real. From here we can simplify our integral by finding $\Re\left(\frac{\sin\left(e^{ix}\right)}{e^{ix}} \right)$.

To avoid clutter, here I defined $c(t) := \cos(t)$ and $s(t):= \sin(t)$. Keeping this in mind, I get that \begin{align} \Re\left(\sin\left(e^{ix}\right)e^{-ix} \right) &= \Re\left(\sin(c + is) (c -is) \right) = \Re\left(\frac{e^{-s}e^{ic}-e^{s}e^{ic}}{2i} (c -is) \right)\\ &=\Re\left(\frac{1}{2}\left(e^{-s}\left[\underbrace{\color{blue}{c\{c\}}}_{\cos(\cos(t)} + i\underbrace{\color{blue}{s\{c\}}}_{\sin(\cos(t)}\right]- e^{s}\left[c\{c\} -i s\{c\}\right] \right) (-s -ic) \right)\\ &=\frac{1}{2} \left(-e^{-s}c\{c\}s + e^{s}c\{c\}s +e^{-s}s\{c\}c +e^s s\{c\}c \right)\\ &=s \cos(c) \left(\frac{e^s -e^{-s}}{2}\right) + c \sin(c) \left(\frac{e^s +e^{-s}}{2}\right)\\ &=\sin(t) \cos(\cos(t))\sinh(\sin(t)) + \cos(t) \sin(\cos(t))\cosh(\sin(t)) \end{align} And here is where I ran into trouble, because I have no idea how I could integrate that last expression. I tried exploiting symmetry, but the function is even, so I don't think I can do much with it without finding an antiderivative (which sounds very unpleasant).

Does anyone know how I could finish my solution? Or alternatively, does anyone know a simpler way in which I can prove this result? Thank you very much!

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3 Answers 3

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Let $z=e^{ix}$. Then the integral becomes

$$\oint_{|z|=1} \frac{\sin(z)}{iz^2}\,dz$$

Can you finish?

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  • $\begingroup$ I think the idea here is the same as in Zachary's answer, but I don't know how I can evaluate $f(0)$ for $f(z) = \frac{\sin(z)}{z}$. $\endgroup$
    – Robert Lee
    Commented Aug 21, 2020 at 0:13
  • $\begingroup$ Good question. Note that $\lim_{z\to0} \frac{\sin(z)}z=1$ $\endgroup$
    – Mark Viola
    Commented Aug 21, 2020 at 0:16
  • $\begingroup$ Ahhh, I think I see. So if I take $$ f(z) = \left\{ \begin{array}{c} \frac{\sin(z)}{z} \quad x \neq 0 \\ 1 \qquad x = 0 \end{array} \right. $$ then I can use this function in Cauchy's formula since it's holomorphic on all $\mathbb{C}$, right? $\endgroup$
    – Robert Lee
    Commented Aug 21, 2020 at 0:22
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    $\begingroup$ Indeed. The singularity is removable. And once removed renders the function entire. $\endgroup$
    – Mark Viola
    Commented Aug 21, 2020 at 0:23
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You should be able to use Cauchy's integral formula. Your integral can be re-written as $$\int_0^{2\pi}f(e^{ix})\,dx,$$ where $f(x)=\sin(x)/x$. Now substitute $u=e^{ix}$, $du/u=idx$ so that your integral becomes $$\frac{1}{i}\int_\gamma \frac{f(u)}{u}\,du.$$ Here, $\gamma$ denotes the unit circle centered at the origin in the complex plane. Cauchy told us that this integral is just $2\pi f(0)$, or in your case, $$2\pi.$$ EDIT: In fact, if $f$ is holomorphic on the unit disk, we have that $$\int_0^{2\pi} f(e^{i\theta})\,d\theta=2\pi f(0).$$

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  • $\begingroup$ This is so much easier, thank you! I just have one question, I would need to verify that $\frac{\sin(z)}{z}$ is holomorphic to apply Cauchy's formula, right? $\endgroup$
    – Robert Lee
    Commented Aug 20, 2020 at 23:46
  • $\begingroup$ I'm also unsure about how we got $f(0) = 1$, wouldn't it be indeterminate since it's $\frac{\sin(0)}{0}$? $\endgroup$
    – Robert Lee
    Commented Aug 20, 2020 at 23:55
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    $\begingroup$ @RobertLee Just the numerator has to be holomorphic on and inside your contour. $\sin(z)$ surely is and we have a pole of order 2. Look at the statement of the formula again. However, I believe there is a mistake in this answer, the final answer should be $2\pi f'(0)$ where $f(z) = sin(z)$. $\endgroup$ Commented Aug 20, 2020 at 23:58
  • $\begingroup$ @NicholasRoberts, I understand that I only need to check $\sin(z)$ to be holomorphic, but I do think that $f(z) = \frac{\sin(z)}{z}$ is correct, since $f(z) = \sin(z)$ doesn't account for the extra $e^{ix}$ you get from the $u-$sub. $\endgroup$
    – Robert Lee
    Commented Aug 21, 2020 at 0:09
  • $\begingroup$ @RobertLee You need to look at the statement of the theorem again. By Mark's answer, we have that our integral is equal to the contour integral of $\int \dfrac{\sin(z)}{iz^2} dz$ over the unit circle. Now you apply the result of Cauchy to the integral $\int \dfrac{f(z)}{iz^2} dz = \frac{1}{i}2\pi if'(0) = 2\pi \cos(0) = 2\pi$ where $f(z) = \sin(z)$. The reason why we are taking the first derivative follows from the $z^2$ in the denominator. Note, I am using the form of the theorem $f^{(n)}(a) =\frac{n!}{2\pi i}\int \frac{f(z)}{(z-a)^{n+1}}$ $\endgroup$ Commented Aug 21, 2020 at 0:24
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Consider the contour integral of $\frac{\sin(z)}{z^2}$ over the circle $\gamma$. Parametrizing the circle over the interval $[-\pi, \pi]$ gives us $i \int \frac{\sin{e^{ix}}}{e^{iz}} dx$.

We can take the Taylor expansion of $\sin(z)$ to get that the contour integral is equal to $\int_\gamma \sum\limits_{i = 0}^\infty \frac{z^{2i - 1}}{(2i + 1)!} dz$. Since the sum is uniformly convergent over the circle, we can swap the sum and the integral to get $\sum\limits_{i = 0}^\infty \int_\gamma \frac{z^{2i -1}}{(2i - 1)!}$. But for $i > 0$, this is the integral of a monomial over a closed path, so the only term that matters is the $i = 0$ term.

Thus, the integral equals $\int_\gamma \frac{1}{z} dz = 2 \pi i$.

Then your original integral is, in fact, $2 \pi$.

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