Question
$$\int_{-\infty}^{+\infty}\frac{\cos x}{\left(1+x+x^2\right)^2+1}\mathrm{~d}x$$
Wolfram alpha says it is
$$\int_{-\infty}^{\infty} \frac{\cos(x)}{\left(1 + x + x^2\right)^2 + 1} \,dx = \frac{\pi (1 + 2\sin(1) + \cos(1))}{5e} \approx 0.745038$$
My try
\begin{align} &\quad\int_{-\infty}^{+\infty}\frac{\cos x}{\left(1+x+x^2\right)^2+1}\mathrm{~d}x\\ &=\underbrace{\int_{-\infty}^{0}\frac{\cos x}{\left(1+x+x^2\right)^2+1}\mathrm{~d}x}_{x\to-x}+\int_{0}^{+\infty}\frac{\cos x}{\left(1+x+x^2\right)^2+1}\mathrm{~d}x\\ &=\int_{0}^{\infty}\left[\frac{\cos x}{\left(1+x+x^2\right)^2+1}+\frac{\cos x}{\left(1-x+x^2\right)^2+1}\right]\mathrm{~d}x\\ &=\int_{0}^{\infty}\frac{\cos x}{\left(1+x^2\right)}\left[\frac{1}{2+2x+x^2}+\frac{1}{2-2x+x^2}\right]\mathrm{~d}x\\ &=2\int_{0}^{\infty}\frac{\left(2+x^2\right)\cos x}{\left(1+x^2\right)\left(4+x^4\right)}\mathrm{~d}x\\ &=\frac{2}{5}\int_0^\infty\frac{\cos x}{1+x^2}\mathrm{~d}x+\frac{12}{5}\int_0^\infty\frac{\cos x}{4+x^4}\mathrm{~d}x\\ &\quad-\frac{2}{5}\int_0^\infty\frac{x^2\cos x}{4+x^4}\mathrm{~d}x\\ &=\frac{2}{5}I_1+\frac{12}{5}I_2-\frac{2}{5}I_3. \end{align}
Alright, now we have three integrals, within which the $I_1$ is a specific case of the so-called $\text{Laplace}$ integral. We can solve it with the following trick.
\begin{align} I_1&=\int_0^\infty\frac{\cos x}{1+x^2}\mathrm{~d}x\\ &=\int_0^\infty\cos x\int_0^\infty\mathrm{e}^{-\left(1+x^2\right)t}\mathrm{~d}t\mathrm{d}x\\ &=\int_0^\infty\mathrm{e}^{-t}\int_0^\infty\mathrm{e}^{-x^2t}\cos x\mathrm{~d}x\mathrm{d}t\\ &=\int_0^\infty\mathrm{e}^{-t}\int_0^\infty\mathrm{e}^{-x^2t}\sum_{n=0}^\infty\frac{(-)^nx^{2n}}{(2n)!}\mathrm{~d}x\mathrm{d}t\\ &=\sum_{n=0}^\infty\frac{(-)^n}{(2n)!}\int_0^\infty\mathrm{e}^{-t}\underbrace{\int_0^\infty\mathrm{e}^{-x^2t}x^{2n}\mathrm{~d}x}_{x^2t\to x}\mathrm{d}t\\ &=\frac{1}{2}\sum_{n=0}^\infty\frac{(-)^n}{(2n)!}\int_0^\infty\mathrm{e}^{-t}t^{-n-\frac{1}{2}}\int_0^\infty\mathrm{e}^{-x}x^{n-\frac{1}{2}}\mathrm{~d}x\mathrm{d}t\\ &=\frac{1}{2}\int_0^\infty\mathrm{e}^{-t}t^{-\frac{1}{2}}\sum_{n=0}^\infty\frac{(-)^n}{(2n)!}\Gamma\left(n+\frac{1}{2}\right)t^{-n}\mathrm{d}t\\ &=\frac{1}{2}\int_0^\infty\mathrm{e}^{-t}t^{-\frac{1}{2}}\sum_{n=0}^\infty\frac{(-)^n(2n-1)!!}{(2n)!2^n}\Gamma\left(\frac{1}{2}\right)t^{-n}\mathrm{d}t\\ &=\frac{\sqrt{\pi}}{2}\int_0^\infty\mathrm{e}^{-t}t^{-\frac{1}{2}}\sum_{n=0}^\infty\frac{1}{n!}\left(-\frac{1}{4t}\right)^n\mathrm{d}t\\ &=\frac{\sqrt{\pi}}{2}\underbrace{\int_0^\infty\mathrm{e}^{-t-\frac{1}{4t}}t^{-\frac{1}{2}}\mathrm{d}t}_{t\to t^2}\\ &=\sqrt{\pi}\int_0^\infty\mathrm{e}^{-t^2-\frac{1}{4t^2}}\mathrm{d}t=\frac{\sqrt{\pi}}{\mathrm{e}}\underbrace{\int_0^\infty\mathrm{e}^{-\left(t-\frac{1}{2t}\right)^2}\mathrm{d}t}_{t-\frac{1}{2t}\to t}\\ &=\frac{\sqrt{\pi}}{\mathrm{e}}\int_{-\infty}^\infty\mathrm{e}^{-t^2}\left(\frac{1}{2}+\frac{t}{2\sqrt{2+t^2}}\right)\mathrm{d}t\\ &=\frac{\sqrt{\pi}}{\mathrm{e}}\int_{0}^\infty\mathrm{e}^{-t^2}\mathrm{d}t\quad (\text{Gauss Integral})\\ &=\frac{\pi}{2\mathrm{e}}. \end{align}
My question is how to evaluate $I_2$ and $I_3$ ? or any other method?
Best Answer
Evaluate $I_2$ and $I_3$ together, i.e. \begin{align} &\int_0^\infty\frac{(6-x^2)\cos x}{4+x^4}dx\\ =& \ \frac12\int_{-\infty}^\infty\frac{(6-x^2)\cos x}{4+x^4}dx\\ =& \ \frac12\int_{-\infty}^\infty\cos x\bigg( \frac{{2x+3}}{\underset{t=x+1}{(x+1)^2+1}}- \frac{{2x+3}}{\underset{t=x-1}{(x-1)^2+1}}\bigg) dx\\ =&\int_{-\infty}^\infty\frac{2t\sin t \sin1+\cos t\cos1}{t^2+1}dt \end{align} where the odd terms resulting from the substitutions $t=x\pm 1$ are discarded because they vanish upon integration. Then, utilize the known integrals $$\int_{-\infty}^\infty \frac{t\sin t}{t^2+1} dt = \int_{-\infty}^\infty \frac{\cos t}{t^2+1} dt= \frac\pi e$$ to arrived at the desired result
$$\int_{-\infty}^{\infty} \frac{\cos x}{\left(1 + x + x^2\right)^2 + 1} \,dx = \frac{\pi }{5e}(1 + 2\sin1 + \cos1) $$