I'm trying to evaluate the integral $$\int_{-\infty}^\infty \frac{\sin (t) \, dt}{t^4+1}$$ using residue and complex plane integration theory.

Let $f(t):=\frac{\sin (t)}{t^4+1}$, $f(z):= \frac{\sin (z)}{z^4+1}$. Then $f(z)$ has four singular points, two of which are in the semicircle $R>2$ in the upper half of the complex plane: $p_1:=\exp\{i\frac{\pi}{4}\}$ and $p_2:=\exp\{i\frac{3\pi}{4}\}$.

We know that $$\int_{-R}^R \frac{\sin (t) \, dt}{t^4+1}+\int_{C_R}\frac{\sin (z) \, dz}{z^4+1}=2\pi i(\operatorname{Res}_{z=p_1}+\operatorname{Res}_{z=p_2})$$

But here's the mysterious part: $$\lim_{R\to\infty}\int_{C_R}\frac{\sin (z) \, dz}{z^4+1}=0$$ yet $$(\operatorname{Res}_{z=p_1}+\operatorname{Res}_{z=p_2})\ne 0$$ But the original integral must be equal to zero. I'd appreciate if it could be pointed out what I'm not doing right.

  • $\begingroup$ Maybe it is useful you include the residues you got. $\endgroup$ – quid Jul 17 '16 at 0:17
  • $\begingroup$ Care to show the computation of residues? Off the top of my head I think they should cancel out (pretty obvious, since $\sin z / (z^4 + 1)$ is an odd function). $\endgroup$ – mathguy Jul 17 '16 at 0:17
  • $\begingroup$ Why $\lim_{R\to\infty}\int_{C_R}\frac{\sin z}{z^4+1}dz=0$? $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$. $\endgroup$ – velut luna Jul 17 '16 at 0:21
  • $\begingroup$ Calculated with Wolfram Alpha: $$Res_{z_0=e^{i\pi/4}} \{f(z)\}=-1/4 (-1)^{1/4} \sin\left((-1)^{1/4}\right)$$ $$Res_{z_0=e^{i3\pi/4}} \{f(z)\}=-1/4 (-1)^{3/4} \sin\left((-1)^{3/4}\right)$$ The sum of these residues is not zero. $\endgroup$ – sequence Jul 17 '16 at 0:26
  • $\begingroup$ @AlphaGo is correct. You need to be thinking about $f(z) = \dfrac{e^{iz}}{z^4+1}$. $\endgroup$ – Ted Shifrin Jul 17 '16 at 0:34

You have $\displaystyle f(t) = \frac{\sin t}{t^4 + 1}$. Instead of considering $f(z)$ for $z \in \Bbb{C}$, consider the function $\displaystyle g(z) = \frac{e^{iz}}{z^4 + 1}$ for $z \in \Bbb{C}$. We do this in general because $e^{iz}$ behaves better (i.e., it's bounded) on $\{z = x + iy \in \Bbb{C} : y \ge 0 \}$ than $\sin z$ does on this set.

Let $\gamma$ be the standard semicircular contour. Then we have $$ \int_\gamma \frac{e^{iz}}{z^4 + 1} \, dz = \int_{-R}^R \frac{e^{it}}{t^4+1} \, dt + \int_{\text{arc}} \frac{e^{iz}}{z^4+1} \, dz,$$

where "arc" represents the curved part of the contour $\gamma$. In can be shown that the integral over "arc" tends to zero as $R \to +\infty$. Just parameterize with $z = Re^{i\theta}$ and use the ML-estimate. Also, we can evaluate the left-hand side with residues: $$ \int_\gamma \frac{e^{iz}}{z^4 + 1} \, dz = 2\pi i \sum_{z \in \Gamma} \text{Res } g(z),$$ where I'm using $\Gamma$ to represent the interior of the closed contour $\gamma$. There are two residues as you mentioned. They occur at $z = e^{\pi i/4}$ and $z = e^{3\pi i/4}$. Note that I'm using exponential form for simplicity. Also note that, in exponential notation, $z^4 + 1$ factors as $$z^4 + 1 = (z-e^{\pi i/4})(z - e^{3\pi i/4})(z - e^{5\pi i/4})(z - e^{7\pi i/4}).$$

First residue: \begin{align} \text{Res}\left(g(z), z= e^{\pi i/4}\right) &= \lim_{z \to e^{\pi i/4}} \left(z - e^{\pi i/4}\right) \frac{e^{iz}}{(z-e^{\pi i/4})(z - e^{3\pi i/4})(z - e^{5\pi i/4})(z - e^{7\pi i/4})}\\[0.3cm] &= \lim_{z \to e^{\pi i/4}} \frac{e^{iz}}{(z - e^{3\pi i/4})(z - e^{5\pi i/4})(z - e^{7\pi i/4})}\\[0.3cm] &= \frac{e^{ie^{\pi i/4}}}{(e^{\pi i/4} - e^{3\pi i/4})(e^{\pi i/4} - e^{5\pi i/4})(e^{\pi i/4} - e^{7\pi i/4})}\\[0.3cm] &= -e^{ie^{\pi i/4}}\frac{\sqrt{2}}{8}(1 + i) \end{align}

Similarly, the other residue is $e^{ie^{3\pi i/4}}\dfrac{\sqrt{2}}{8}\left(1 - i\right)$. Yes, I'm skipping a lot of messy arithmetic here. Exercise left for the reader. :)

Anyway, with the help of a computer algebra system, I find that the sum of the residues is (equivalent to) $$ -\frac{i}{4} e^{-1/\sqrt{2}}\sqrt{2}\left(\cos \frac{1}{\sqrt{2}} + \sin\frac{1}{\sqrt{2}}\right). $$ This is a mess. But that's ok. Note that it's purely imaginary, and that's the only thing we care about regarding this expression. This means that if we multiply it by $2\pi i$, we get a real value. This means that $$ \int_\gamma \frac{e^{iz}}{z^4 + 1} \, dz $$ is real. Finally, note also that $e^{it} = \cos t + i\sin t$. Therefore, when we take the limit as $R \to +\infty$, we have: \begin{align} \int_\gamma \frac{e^{iz}}{z^4 + 1} \, dz &= \int_{-\infty}^{+\infty} \frac{e^{it}}{t^4+1} \,dt\\[0.3cm] &= \int_{-\infty}^{+\infty} \frac{\cos t + i\sin t}{t^4 + 1} \, dt\\[0.3cm] &= \underbrace{\int_{-\infty}^{+\infty} \frac{\cos t}{t^4+1} \, dt}_{\text{real part}} + i \cdot \underbrace{\int_{-\infty}^{+\infty} \frac{\sin t}{t^4 + 1} \, dt}_{\text{imaginary part}} \end{align}

Recall that the integral over $\gamma$ is real. Therefore its imaginary part is zero. So if we equate real and imaginary parts then we get $$ \int_{-\infty}^{+\infty} \frac{\sin t}{t^4 + 1} \, dt = 0.$$


This :$$\lim_{R\to\infty}\int_{C_R}\frac{\sin (z) \, dz}{z^4+1}=0$$ is false because of the sine. (Which is NOT bounded on $\mathbb{C}$) The usual trick is to consider the integral $$\int_{-\infty}^{+\infty} \frac{e^{it}}{t^4+1} dt,$$ to compute it by residue using Jordan's Lemma and then to take the imaginary part to get the integral you want.


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