Respuesta :
I use complex analysis to compute the integrals in question.
First, notice that the first integrand is even:
[tex]\dfrac{\cos(-x)}{(-x)^2+1}=\dfrac{\cos(x)}{x^2+1}[/tex]
[tex]\implies\displaystyle\int_0^\infty\frac{\cos(x)}{x^2+1}\,dx=\frac12\int_{-\infty}^\infty\frac{\cos(x)}{x^2+1}\,dx[/tex]
Consider a contour C that's the union of
• Γ, a semicircle of radius R in the upper half-plane, and
• the line segment connecting the points (-R, 0) and (R, 0)
On Γ, we have [tex]z=Re^{it}[/tex] with 0 ≤ t ≤ π.
Consider the complex function
[tex]f(z)=\dfrac{e^{iz}}{z^2+1}[/tex]
and notice that our original integrand is the real part of f(z). Then the integral of f(z) over C is
[tex]\displaystyle\int_Cf(z)\,dz=\lim_{R\to\infty}\left(\int_{-R}^Rf(z)\,dz+\int_\Gamma f(z)\,dz\right)[/tex]
As R → ∞, the first integral on the right is exactly twice the one we want. Estimate the second one to be bounded by
[tex]\displaystyle\left|\int_\Gamma f(z)\,dz\right|\le\pi R|f(z)|\le\frac{\pi R}{R^2-1}[/tex]
since
[tex]|z^2+1|\ge\bigg||z^2|-|-1|\bigg|=|R^2-1|[/tex]
and so the integral along Γ vanishes.
f(z) has only one pole in the interior of C at z = i. By the residue theorem,
[tex]\displaystyle\int_Cf(z)\,dz=2\pi i\,\mathrm{Res}\left(f(z),z=i\right)=2\pi i\lim_{z\to i}(z-i)f(z)=\frac\pi e[/tex]
[tex]\implies\displaystyle\int_0^\infty\frac{\cos(x)}{x^2+1}\,dx=\frac12\mathrm{Re}\left(\int_Cf(z)\,dz\right)=\frac\pi{2e}[/tex]
For the second integral, we recall that for complex z,
[tex]\sqrt z=\exp\left(\dfrac12\left(\ln|z|+i\arg(z)\right)\right)[/tex]
Consider a keyhole contour C, the union of
• [tex]\Gamma_R[/tex], the larger circle with radius R and [tex]z=Re^{it}[/tex], with 0 < t < 2π ;
• [tex]\Gamma_\varepsilon[/tex], the smaller circle with radius ε and [tex]z=\varepsilon e^{-it}[/tex], with 0 < t < 2π ;
• [tex]\ell_1[/tex], the line segment above the positive real axis joining [tex]\Gamma_\varepsilon[/tex] to [tex]\Gamma_R[/tex] ; and
• [tex]\ell_2[/tex], the other line segment below the positive real axis joining [tex]\Gamma_R[/tex] to [tex]\Gamma_\varepsilon[/tex]
Then
[tex]\displaystyle\int_Cf(z)\,dz=\int_{\Gamma_R}f(z)\,dz+\int_{\ell_1}f(z)\,dz+\int_{\Gamma_\varepsilon}f(z)\,dz+\int_{\ell_2}f(z)\,dz[/tex]
and in the limit, the integral over [tex]\ell_1[/tex] converges to the one we want.
Estimate the integrals over the circular arcs:
• [tex]\Gamma_R[/tex] :
[tex]\displaystyle\left|\int_{\Gamma_R}f(z)\,dz\right|\le2\pi R|f(Re^{it})|\le\dfrac{2\pi R^{3/2}}{|R-\sqrt5|^2}\to0[/tex]
as R → ∞.
• [tex]\Gamma_\varepsilon[/tex] :
[tex]\displaystyle\left|\int_{\Gamma_\varepsilon}f(z)\,dz\right|\le2\pi \varepsilon|f(\varepsilon e^{-it})|\le\dfrac{2\pi\varepsilon^{3/2}}{|\varepsilon-\sqrt5|^2}\to0[/tex]
as ε → 0.
Consider the integral over [tex]\ell_2[/tex] :
[tex]\displaystyle\int_{\ell_2}f(z)\,dz=\int_R^\varepsilon\frac{\sqrt z}{z^2+2z+5}\,dz\\\\=\int_R^\varepsilon\frac{\exp\left(\dfrac12\left(\ln|z|+2\pi i\right)\right)}{z^2+2z+5}\,dz\\\\=-\int_R^\varepsilon\frac{\exp\left(\dfrac12\ln|z|\right)}{z^2+2z+5}\,dz\\\\=\int_\varepsilon^R\frac{\sqrt z}{z^2+2z+5}\,dz\\\\=\int_{\ell_1}f(z)\,dz[/tex]
so in fact,
[tex]\displaystyle\int_Cf(z)\,dz=2\int_0^\infty\frac{\sqrt x}{x^2+2x+5}\,dx[/tex]
By the residue theorem,
[tex]\displaystyle\int_Cf(z)\,dz=2\pi i\sum_{\rm poles}\mathrm{Res}\,f(z)[/tex]
We have poles at z = -1 + 2i and z = -1 - 2i. On our chosen branch,
[tex]\sqrt{-1+2i}=i\sqrt[4]{5}\exp\left(-\dfrac i2\tan^{-1}(2)\right)[/tex]
[tex]\sqrt{-1-2i}=i\sqrt[4]{5}\exp\left(\dfrac i2\tan^{-1}(2)\right)[/tex]
The residues are
[tex]\mathrm{Res}(f(z),z=-1-2i)=\dfrac{i\sqrt[4]{5}\exp\left(\frac i2\tan^{-1}(2)\right)}{-4i}[/tex]
[tex]\mathrm{Res}(f(z),z=-1+2i)=\dfrac{i\sqrt[4]{5}\exp\left(-\frac i2\tan^{-1}(2)\right)}{4i}[/tex]
Their sum is
[tex]\displaystyle\sum_{\rm poles}\mathrm{Res}\,f(z)=-\frac{\sqrt[4]{5}}2\sin\left(\dfrac12\tan^{-1}(2)\right)=-\frac{\sqrt[4]{5}}2\sqrt{\frac{5-\sqrt5}{10}}=-\frac i2\sqrt{\frac1\phi}[/tex]
where ɸ = (√5 + 1)/2 is the golden ratio, and so the overall integral is
[tex]\displaystyle\int_0^\infty\frac{\sqrt x}{x^2+2x+5}\,dx=\frac\pi2\sqrt{\frac1\phi}[/tex]
Lastly, recall
[tex]\displaystyle\sum_{k=0}^\infty\frac1{k!}=e[/tex]
Then our expression reduces to
[tex]\left(\dfrac{e\times\frac\pi{2e}}{\frac\pi2\sqrt{\frac1\phi}}\right)^2=\boxed{\phi}[/tex]
