I'm considering here the fact that $$\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {e^{iz}}{z^2+1} dz=0$$ , where $\Gamma$ is a contour defined as a semicircle centred about the origin, of radius $R>1$, in the upper half-plane of $\mathbb{C}$, positively oriented.

Now, I think, one could express $\int\limits_{-\infty}^{\infty} \frac {sin(t)}{t^2+1} dt$ as $$\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {sin(t)}{t^2+1} dt$$ But here's where I don't know what to do next. Would appreciate a hint.

On a side note, $$\int\limits_{-\infty}^\infty \frac {e^{iz}}{z^2+1} dz\ne 0$$ , so I'm not sure what I'm not seeing here because $$\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {e^{iz}}{z^2+1} dz=0$$, so where's the analogy?

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    $\begingroup$ Are you aware of the residue theorem? $\endgroup$ – joriki Jun 16 '16 at 23:35
  • $\begingroup$ Haven't studied it yet. $\endgroup$ – sequence Jun 17 '16 at 0:03

The integrand is and odd function of $t$ and the integral is absolutely convergent, so $$\int_{-\infty}^{\infty}\frac{\sin t}{t^2+1}dt=0$$ EDIT: Writing out the theorem longhand, $$\begin{align}\int_{-\infty}^{\infty}\frac{\sin t}{t^2+1}dt&=\lim_{M\rightarrow-\infty}\int_M^0\frac{\sin t}{t^2+1}dt+\lim_{N\rightarrow\infty}\int_0^N\frac{\sin t}{t^2+1}dt\\ &=\lim_{M\rightarrow-\infty}\int_{-M}^0\frac{\sin(-u)}{u^2+1}(-du)+\lim_{N\rightarrow\infty}\int_0^N\frac{\sin u}{u^2+1}du\\ &=\lim_{M\rightarrow-\infty}-\int_0^{-M}\frac{\sin u}{u^2+1}du+\lim_{N\rightarrow\infty}\int_0^N\frac{\sin u}{u^2+1}du\\ &=-\int_0^{\infty}\frac{\sin u}{u^2+1}du+\int_0^{\infty}\frac{\sin u}{u^2+1}du=0\end{align}$$ Where we have used the substitution $u=-t$ in the left integral and $u=t$ in the right integral. Notice that all that was required for this proof to work was that the integrand was an odd function, the interval of integration was $(-\infty,\infty)$, although an interval of $(-a,a)$ would have worked and that the function was Riemann integrable over the interval. This theorem can reduce some perfectly horrible-looking integrals to zero with relatively little effort.

  • $\begingroup$ Can you please be more specific? $\endgroup$ – sequence Jun 17 '16 at 0:04
  • $\begingroup$ @sequence: if $f\in L^1(\mathbb{R})$ and $f$ is an odd function, $$\int_{-\infty}^{+\infty}f(x)\,dx = \color{red}{0}.$$ $\endgroup$ – Jack D'Aurizio Jun 17 '16 at 0:14
  • $\begingroup$ What is $L^1(\mathbb{R})$? $\endgroup$ – sequence Jun 17 '16 at 0:32

First Solution:

From Residue theorem: $$\int\limits_{-\infty}^\infty \frac {e^{iz}}{z^2+1} dz = \frac{\pi}{e}$$

By Euler's formula: $$\int\limits_{-\infty}^\infty \frac {e^{iz}}{z^2+1} dz =\int\limits_{-\infty}^\infty \frac{cos z}{z^2+1} dz + i\int\limits_{-\infty}^\infty \frac{sin z}{z^2+1} dz$$

In this post, it shows that: $$\int\limits_{-\infty}^\infty \frac{\cos z}{z^2+1} dz = \frac{\pi}{e}$$

In conclude: $$\int\limits_{-\infty}^\infty \frac{\sin z}{z^2+1} dz = 0$$

Alternative Solution(inspired by the same post):

Because $sin(z)$ behaves not like $e^{iz}$ over the complex plane, the straightforward way to use residue theorem is invalid.

By Euler's formula: $$\int\limits_{-\infty}^\infty \frac {\sin z}{z^2+1} dz = \frac{1}{2} i (\int\limits_{-\infty}^\infty \frac {e^{-ix}}{z^2+1} dz-\int\limits_{-\infty}^\infty \frac {e^{ix}}{z^2+1} dz) $$

From Residue theorem: $$\int\limits_{-\infty}^\infty \frac {e^{ix}}{z^2+1} dz = \frac{\pi}{e} $$ $$\int\limits_{-\infty}^\infty \frac {e^{-ix}}{z^2+1} dz = \frac{\pi}{e} $$

The proof is done.

  • $\begingroup$ The Residue Theorem has not been studied yet, so a solution should be possible without it. $\endgroup$ – sequence Jun 17 '16 at 0:32
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    $\begingroup$ That doesnt matter. The key issue about your analogy is that $\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {sin(t)}{t^2+1} dt \neq 0$ even though $\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {e^{iz}}{z^2+1} dz=0$ $\endgroup$ – Zack Ni Jun 17 '16 at 0:36
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    $\begingroup$ The Residue Theorem is a very simple theorem about the relation between poles and closed contour integration. Hopefully, you can learn it in several secs. $\endgroup$ – Zack Ni Jun 17 '16 at 0:38

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