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In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $\text{Proposition (1)}$ is sound ?

$\text{Proposition (1)}$

$$\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$$

$\text{Proof}$

Assume that $R>1$ define $\gamma_{R}$ such that,

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

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

$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ enter image description here

Consider our choice $f$ and that,

$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz.$$

Clearly it's obvious that

$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3}dz = \sum_{\psi = 1,2} \oint_{\gamma_{R}^{\psi}}\frac{e^{iz}}{z^{2}+3}dz.$$

It's trivial that,

$$\oint_{\gamma_{R}^{1}}e^{iz}/({z^{2}+3})dx \rightarrow \int_{0}^{\infty} \frac{e^{ix}}{x^{2}+3}\operatorname{dx}$$

It's natural to claim that,

$$\Bigg| \lim_{R \rightarrow \infty}\oint_{ \gamma_{R}^{2}} \frac{e^{iz}}{z^{2}+3} dz \Bigg| \rightarrow 0. $$

Using the Estimation Lemma one can be relived that,

$$\bigg |\oint_{\gamma_{R}^{2}} \frac{e^{iz}}{z^{2}+3} dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|\frac{e^{iz}}{z^{2}+3}|\leq \pi R \cdot \frac{1}{R^{2} - 3} \rightarrow 0 \, \text{as} \, R \rightarrow \infty$$

Thus,

$$ \operatorname{Re}\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \operatorname{Re} \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}} = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}.$$

However we need to consider that

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

It's easy to observe that,

$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz = \big(2\pi i \cdot \frac{ie^{\sqrt{3}}}{2 \sqrt{3}}) = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$$

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    $\begingroup$ Where does $\frac1{R^2-1}$ come from? $\endgroup$ Aug 7, 2018 at 18:41
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    $\begingroup$ Thanks for pointing out the typo it's been fixed. $\endgroup$
    – Zophikel
    Aug 7, 2018 at 18:47
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    $\begingroup$ The integral is over $\Bbb R$, or only over $\Bbb R_{\ge 0}$? (The contour goes from $-R$ to $R$. I was looking for a factor $2$ or $1/2$ on the one or the other side...) $\endgroup$
    – dan_fulea
    Aug 7, 2018 at 19:14
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    $\begingroup$ More generally, $$\int_0^\infty\,\frac{\cos(kx)}{x^2+1}\,\text{d}x=\frac{\pi}{2}\,\exp\big(-|k|\big)\text{ for all }k\in\mathbb{R}\,.$$ In your case, $$\int_0^\infty\,\frac{\cos(x)}{x^2+3}\,\text{d}x=\frac{1}{\sqrt{3}}\,\int_0^\infty\,\frac{\cos(\sqrt{3}t)}{t^2+1}\,\text{d}t=\frac{\pi}{2\sqrt{3}}\,\exp(-\sqrt{3})\,,$$ where $t:=\sqrt{3}x$. See, for example, the last two hidden boxes in the question of math.stackexchange.com/questions/2870410/…. $\endgroup$ Aug 7, 2018 at 19:37
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    $\begingroup$ My comment is that there are two errors. Note that $$\lim_{R\to\infty}\,\int_{\gamma_R^1}\,\frac{\exp(\text{i}z)}{z^2+3}\,\text{d}z=2\,\int_0^\infty\,\frac{\cos(x)}{x^2+3}\,\text{d}x$$ as you probably forgot that $2\,\cos(x)=\exp(+\text{i}x)+\exp(-\text{i}x)$, and that $$2\pi\text{i}\cdot\left(\frac{\exp(-\sqrt{3})}{2\sqrt{3}\text{i}}\right)=\frac{\pi\,\exp(-\sqrt{3})}{\sqrt{3}}$$ is what you should get after applying the Residue Theorem. Dividing this result by $2$, you will get the answer you are looking for. $\endgroup$ Aug 8, 2018 at 7:50

1 Answer 1

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Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.

$\text{Proof}$

To proceed one must consider that,

$$ \int_0^\infty\,\frac{\cos(x)}{x^2+3}\,\text{d}x=\frac{1}{\sqrt{3}}\,\int_0^\infty\,\frac{\cos(\sqrt{3}t)}{t^2+1}\,\text{d}t=\frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}$$

$t:=\sqrt{3}x$

Assume that $R>1$ define $\gamma_{R}$ such that,

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

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

Consider our choice $f$ and that,

$$\frac{1}{\sqrt{3}}\,\oint_{\gamma_{R}}\,\frac{e^{(\sqrt{3}t)}}{z^2+1}\,\text{d}z $$

Clearly it's simple that $$\frac{1}{\sqrt{3}} \oint_{\gamma_{R}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz = \sum_{\psi = 1,2} \frac{1}{\sqrt{3}} \oint_{\gamma_{R}^{\psi}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz.$$

Now it's imperative to claim that

$$\lim_{R\to\infty}\,\oint_{\gamma_R^1}\,\frac{\exp(\sqrt{3}z)}{z^2+3}\,\text{d}z= \frac{1}{\sqrt{3}} \,\int_{0}^{\infty}\, \, \frac{e^{(\sqrt{3}t)}}{z^2+1}\,\text{d}z = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}.$$

Now at this leg of our journey important to conjecture that,

$$\Bigg| \lim_{R \rightarrow \infty}\oint_{ \gamma_{R}^{2}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz \Bigg| \rightarrow 0. $$

Thanks to Estimation Lemma we can be relived that,

$$\bigg |\oint_{\gamma_{R}^{2}} \frac{e^{(\sqrt{3}z)}}{z^2+3}dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|\frac{e^{(\sqrt{3}z)}}{z^2+3}|\leq \pi R \cdot \frac{1}{R^{2} - 3} \rightarrow 0 \, \text{as} \, R \rightarrow \infty.$$

Thus, taking

$$\operatorname{Re} \bigg( \frac{1}{\sqrt{3}}\,\int_0^\infty\,\frac{\cos(\sqrt{3}t)}{t^2+1}\text{d}t \bigg) \, = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}$$

However we need to consider that

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

After a trivial calculation we have that,

$$\frac{1}{\sqrt{3}} \oint_{\gamma_{R}} \frac{e^{(\sqrt{3}z)}}{z^2+1}\, dz = 2 \pi i + \operatorname{Res}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}.$$

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    $\begingroup$ @Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ? $\endgroup$
    – Zophikel
    Aug 8, 2018 at 15:38
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    $\begingroup$ Rather than posting your (additional) work as an answer, better to edit it into the original post. $\endgroup$
    – Doug M
    Aug 17, 2018 at 0:58
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    $\begingroup$ Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski $\endgroup$
    – Zophikel
    Aug 17, 2018 at 1:44

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