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let's define $\begin{array}{c} f=\begin{cases} \frac{\sin x}{x} & x\neq0\\ 1 & x=0 \end{cases}\end{array}$ f is holomorphic on $\mathbb{C}$ as it equals to it's taylor series. therefore, for each domain, according the residue theorem $\oint_{c}f$ it should be exactly 0, as there are no poles in any domain. Where am I wrong. of course this result cannot be true as it implies that $\oint_{c}\frac{\sin x}{x}dx=0$ because they differ by a single point. this of course cannot be true as it implies $\intop_{0}^{\infty}\frac{\sin x}{x}dx=0\neq\frac{\pi}{2}$

hash man
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    $\int_0^\infty f(x),dx$ is not an integral over a closed contour. – Angina Seng Aug 02 '20 at 09:16
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    Your error is more evident if you observe that the only thing you are using of $f$ is that is holomorphic on $\mathbb{C}$. Every entire function $f$ (like a polynomial or $e^z$) satisfies $\oint_C f(z) = 0$ for every CLOSED contour. – jjagmath Aug 02 '20 at 09:25
  • @AnginaSeng yeah but I can convert it to half a circle, which is made up from an arch integral that tends to 0, and -R,R integral – hash man Aug 02 '20 at 09:53
  • @hashman Why do you think the arch integral tends to $0$? It goes over area where $\sin z$ can grow quite a lot in modulus. $\sin iz=i\sinh z$ IIRC. –  Aug 02 '20 at 10:38
  • Things people know from real analysis: sine is a bounded function. Things people learn in complex analysis: sine is an entire function, and bounded entire functions are constant. Conclusion: sine is constant? – Sten Aug 02 '20 at 11:15

3 Answers3

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The OP has clarified in comments that they are thinking about taking the limit of $\oint_c f(z)\,dz$ where the closed contour $C$ is a half circle running from $-R$ to $+R$ and then back to $-R$ along the arc $z=Re^{i\theta}$ with $0\le\theta\le\pi$, as $R\to\infty$. They are correct that this limit is $0$, since the function $f(z)$ is holomorphic. But all this proves is that

$$\int_0^{\pi}{\sin(Re^{i\theta})\over Re^{i\theta}}Rie^{i\theta}\,d\theta=i\int_0^{\pi}\sin(Re^{i\theta})\,d\theta=-\int_{-R}^R{\sin x\over x}dx$$

hence

$$\lim_{R\to\infty}\int_0^{\pi}\sin(Re^{i\theta})\,d\theta=i\pi$$

The OP's error is in assuming that the integral over the semicircular arc tends to $0$ just because the arc is heading off to infinity and $\sin z\over z$ looks like it's getting small as $|z|$ gets large; in point of fact, for $z=iy$ (with $y$ real) we have $\left|\sin z\over z\right|={\sinh y\over y}\to\infty$ as $|y|\to\infty$, so the function is anything but small on portions of any large semicircular arc.

Barry Cipra
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  • actually isn't there are problem that 0 is on the bondary? – hash man Aug 02 '20 at 11:37
  • still even if we look at the original proof: https://math.stackexchange.com/a/601898/799836 now we have a regular point, which means that epsilon circle is now then evaluates to 0 and not $-\pi i$ – hash man Aug 02 '20 at 11:41
  • The question you linked to is trying to evaluate the integral of $\sin x\over x$ by doing a clever contour integral with the function $e^{iz}/z$. That function does have a pole at $z=0$, so it's necessary for the contour to have an epsilon circle; the function $\sin z\over z$ does not have a pole at $z=0$, so there's no need for any epsilonish arcs. – Barry Cipra Aug 02 '20 at 13:42
  • okay so why putting instead of evaluating $e^{iz}/z$ don't we evaluate $sin(z)i/z$ on the complex plane and don't get the 0 result as an answer? – hash man Aug 02 '20 at 13:53
  • @hashman, sorry, I've explained things about as well as I feel capable of. Hopefully someone else will post an answer that makes complete sense to you. Or, perhaps, you'll be able to reread this later when your understanding of the subject deepens. If you're taking complex variables as a formal course, I recommend putting your tuition dollars to work, and have your instructor explains things. – Barry Cipra Aug 02 '20 at 16:08
  • you want to claim that sin also has a pole on the domain? this is it? – hash man Aug 02 '20 at 16:14
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The error is in the last part. $\oint_C f(z) dz=0$ doesn't implies $\int_0^\infty f(z) dz = 0$.

jjagmath
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I hope your function $f$ is a complex variable function $f(z)$ so that holomorphic makes some sense here.

The result $\oint_C f(z)dz=0$ is obviously true for some close contour $C$ like the union of real axis and semicircular arc $\lvert z\lvert =1$ which doesn't imply that $\oint_C f(x)dx=0$ unless $f(x)=\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials.

Nitin Uniyal
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