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Where is the error in evaluating this integration:$$\int^{\infty}_{-\infty}\frac{\cos(2x)}{(1+x^2)^2}dx$$ enter image description here

we have: $$\int_{C_R}\frac{\cos(2z)}{(1+z^2)^2}dz=\int_{-R}^{R}\frac{\cos(2x)}{(1+x^2)^2}dx+\int_{\Gamma_R}\frac{\cos(2z)}{(1+z^2)^2}dz=2i\pi\text{Res}(f(z),z=i)=\frac{\pi(3-e^4)}{2e^2}$$

and we have : $$0\le | \int_{\Gamma_R}\frac{\cos(2z)}{(1+z^2)^2}dz | \le \frac{\pi R}{(R^2-1)^2}$$ Let $$R\to \infty $$ this implies that :$$\int_{\Gamma_R}\frac{\cos(2z)}{(1+z^2)^2}dz=0$$

therfore: $$\int_{-\infty}^{\infty}\frac{\cos(2x)}{(1+x^2)^2}dx=\frac{\pi(3-e^4)}{2e^2}$$ but this is result is wrong Because using wolfram alpha : $$\int_{-\infty}^{\infty}\frac{\cos(2x)}{(1+x^2)^2}dx=\frac{3\pi}{2e^2}$$

Delta
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1 Answers1

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You noticed that $\left|\cos z\right|\leq 1$ was a mistake. If this held throughout $\mathbb C$, then $\cos(z)$ would be constant by Liouville's theorem. This led to the incorrect estimate

$$\left|\int_{\Gamma_R} \frac{\cos(2z)}{(1+z^2)^2}\right|\leq \frac{\pi R}{(R^2 -1)^2}$$

which I assume came about as follows:

\begin{align*} \left|\int_{\Gamma_R}\frac{\cos(2z)}{(1+z^2)^2}dz\right| &= \left|\int_0^\pi\frac{\cos(2\cdot Re^{i\theta})}{(1+( Re^{i\theta})^2)^2}\cdot iRe^{i\theta}d\theta\right|\\ &\leq \int_0^\pi\frac{\left|\cos(2Re^{i\theta})\right|}{\left|1+( Re^{i\theta})^2\right|^2}\cdot \left|iRe^{i\theta}\right|d\theta\\ &= R\int_0^\pi\frac{\left|\cos(2Re^{i\theta})\right|}{\left|1+( Re^{i\theta})^2\right|^2}d\theta &\text{(since $\left|iRe^{i\theta}\right|=R$)}\\ &\leq R\int_0^\pi\frac{1}{\left|1+( Re^{i\theta})^2\right|^2}d\theta &{\color{red}{\text{(since $\left|\cos x\right|\leq 1$)}}}\\ &\leq \cdots \end{align*}

The correct bound would be $\left|\cos(2Re^{i\theta})\right|\leq \left|\cos(2Re^{i\pi /2})\right|=\cosh(2R)$ as $\cos z$ exhibits its maximal growth rate along the imaginary axis, leading to $$\left|\int_{\Gamma_R} \frac{\cos(2z)}{(1+z^2)^2}\right|\leq \frac{\pi R}{(R^2 -1)^2}\cosh(2R)$$ which is not helpful as the right side is unbounded as $R\to\infty$.

Alann Rosas
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