Calculate using the residues theorem this integral : $$ \int \limits_{-\infty}^{+\infty} \frac{\mathrm{d}x}{x^4+1}. $$

First I calculated $\displaystyle \int_{C_r} \frac{\mathrm{d}z}{z^4+1} $, where $C_r$ is a half circle centered in the origin and it radius $r>1$ including the real axis, I found (by summing the residues) that it equals : $\displaystyle \frac{\pi}{\sqrt{2}}$, considering : $T_r$ as the arc of $C_r$ (no real axis) I got : $$ \frac{\pi}{\sqrt{2}} = \int\limits_{-r}^{r} \frac{\mathrm{d}z}{z^4+1} + \int_{T_r}\frac{\mathrm{d}z}{z^4+1} .$$

Now, what can we say about $$\lim_{r\to \infty} \int_{T_r}\frac{\mathrm{d}z}{z^4+1}$$ It seem that it tends toward zero but I cant prove it. if you think that it is easy, then just post hints.

Some googling lead me to this, how do we prove it ? (the easiest method).

Thanks in advance.

  • $\begingroup$ What is $T_r$?${}$ $\endgroup$ – Git Gud Apr 14 '13 at 16:41
  • $\begingroup$ presumably the arc... @GitGud $re^{i\theta}$ for $\theta$ from 0 to $\pi$ $\endgroup$ – Lost1 Apr 14 '13 at 16:42
  • $\begingroup$ it is the arc : $z\in T_r$ means that $ \text{Im}(z)>0$ and $|z|=r$. $\endgroup$ – aziiri Apr 14 '13 at 16:43

The points of $T_r$ are of the form $z(\theta) = r\operatorname{e}^{i\theta}$ where $0 \le \theta \le \pi$. We can substitute this into the integral. Forst notice that $\operatorname{d}\!z = ir\operatorname{e}^{i\theta}\, \operatorname{d}\!\theta$, giving:

$$\int_{T_r} \frac{1}{z^4+1} \, \operatorname{d}\!z = \int_0^{\pi} \frac{ir\operatorname{e}^{i\theta}}{r^4\operatorname{e}^{4i\theta}+1} \, \operatorname{d}\!\theta $$

We can estimate the size of this integral:

\begin{array}{ccc} \left|\int_0^{\pi} \frac{ir\operatorname{e}^{i\theta}}{r^4\operatorname{e}^{4i\theta}+1} \, \operatorname{d}\!\theta\right| &\le& \int_0^{\pi} \left|\frac{ir\operatorname{e}^{i\theta}}{r^4\operatorname{e}^{4i\theta}+1} \right| \operatorname{d}\!\theta \\ &=& \int_0^{\pi}\frac{r}{\left|r^4\operatorname{e}^{4i\theta}+1\right|} \operatorname{d}\!\theta \\ &\le& \int_0^{\pi}\frac{r}{\left|r^4\operatorname{e}^{4i\theta}\right|-\left|1\right|} \operatorname{d}\!\theta \\ &=& \int_0^{\pi}\frac{r}{r^4-1} \operatorname{d}\!\theta \end{array} In step three I applied a version of the triangle inequality which says that, for all $z,w \in \mathbb{C}$ we have $|z+w| \ge ||z|-|w||$. As $r \to \infty$ this last integral tends towards zero. Hence our original internal terns to zero as $r \to \infty$.


We have that

$$\text{On}\;\;C_r\,,\;\;|z|=r\implies \left|\;\int\limits_{C_r}\frac{1}{z^4+1}dz\;\right|\le\frac{1}{r^4-1}\pi r\xrightarrow[r\to\infty]{}0$$

  • $\begingroup$ thanks, this consist on the estimation lemma which I mentioned in the last line, how do we prove it ? $\endgroup$ – aziiri Apr 14 '13 at 16:49
  • $\begingroup$ Wow, how lucky I am you didn't give a -1...phew! Anyway, the OP said he already googled the estimation lemma so if he can't manage after reading that article I think he may use a whole written example, and besides: on this any hint is practically a solution, imfho. $\endgroup$ – DonAntonio Apr 14 '13 at 16:51
  • $\begingroup$ How would we prove it? Parameterize the semi-circle, then use real-variable estimates for an integral. $\endgroup$ – GEdgar Apr 14 '13 at 16:52
  • $\begingroup$ @aziiri, how do we prove what? The estimation lemma? It follows, I believe, from the parallel estimation of (real) line integrals. $\endgroup$ – DonAntonio Apr 14 '13 at 16:52
  • $\begingroup$ Okay, I got it now. $\endgroup$ – aziiri Apr 14 '13 at 16:54

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