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In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?

$\text{Proposition} \, \, \, (1) $

$$\int_{-\infty}^{\infty}\left( \frac{\cos{\left (x \right )}}{x^{4} + 1} \right)dx=\frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right) $$

The first step on our quest to prove $(1)$, is one must consider a Semicircular Contour $\gamma_{R}$, assuming $R > 1$ define

$$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R $$

$$\gamma_{R}^{2}(t) = Re^{it} \, \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$

One can call these two curves taken together $\gamma_{R}$ or $\gamma$, after picking our $\gamma_{R}$ one can consider $$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz.$$

It's trivial that,

$$\oint_{\gamma_{R}}g(z)~ dz = \oint_{\gamma_{R}^{1}} g(z) dz+ \oint_{\gamma_{R}^{2}}g(z) ~dz.$$

It's imperative that

$$\displaystyle \oint_{\gamma_{R}^{1}} g(z) dz \rightarrow \lim_{R \rightarrow \infty }\int_{-R}^{R} \frac{e^{ix}}{1+x^{4}} \, \, \text{as}\, \, R \rightarrow \infty $$

Using the Estimation Lemma one be relived to see

$$\bigg |\oint_{\gamma_{R}^{2}} g(z) dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|g(z)|\leq \pi \cdot \frac{1}{R^{4} - 1}. $$

Now it's safe to say that

$$\lim_{R \rightarrow \infty}\bigg | \oint_{\gamma_{R}^{2}}\frac{1}{1+z^{4}} dz \bigg| \rightarrow 0. $$

It's easy to note after all our struggle that

$$ \int_{-\infty}^{\infty} \frac{cos(x)}{1+x^{4}} = Re \int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^{4}} = Re\bigg( \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right)\bigg) = \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right) $$

The finial leg of our conquest is to consider that

$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i \sum_{j=1,2,3,4} Ind_{\gamma} \cdot \operatorname{Res_{f}(P_{j})}$$

It's easy to calculate that

$$\operatorname{Res_{z = \sqrt[4]{-1}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}}$$

$$\operatorname{Res_{z = \sqrt[4]{-1}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}}$$

$$\operatorname{Res_{z = -(-1)^{3/4}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}$$

$$\operatorname{Res_{z = (-1)^{3/4}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}$$

Putting the pieces together we have that

$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i + \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}= \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right).$$

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  • $\begingroup$ Note that $\sin(x)+\cos(x) = \sqrt{2}\sin(x+\pi/4)$. $\endgroup$ Aug 5, 2018 at 23:37
  • $\begingroup$ If you are interested in any alternative approach to the question, please advise and I will submit. $\endgroup$
    – user150203
    Dec 10, 2018 at 23:26
  • $\begingroup$ Sure @DavidG I'd like a hint to an alternate way to solve this problem :>). $\endgroup$
    – Zophikel
    Dec 11, 2018 at 0:57

2 Answers 2

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Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $\exp\left(\frac{\pi i}4\right)$, $\exp\left(\frac{3\pi i}4\right)$, $\exp\left(\frac{5\pi i}4\right)$, and $\exp\left(\frac{7\pi i}4\right)$. But only the first two matter, since they're the only ones which are in the semicircle bounded by the image of $\gamma_R$.

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  • $\begingroup$ Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $\frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}}$ and $\frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}}$. I'll have answer my question with the correct final steps where I apply the Residue Theorem $\endgroup$
    – Zophikel
    Aug 5, 2018 at 23:42
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From the previous attempt the last line,

$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i + \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}= \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right).$$

Is incorrect and one should sum the poles inside $\gamma_{R}$ hence,

$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i +\operatorname{Res}\limits\limits_{ z=\omega}\frac{e^{iz}}{z^4+1} = \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right). $$

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