Question

`\left \lgroup\displaystyle\rm \sum_(k=0)^(\infty){1\over k!}\int_0^(\infty){\cos(x)\over \displaystyle x^2+1}\text{d}x \over \displaystyle \rm \int_0^\infty{√(x)\over x^2+2x+5}dx \right \rgroup^(2)`​

Answer 1

I use complex analysis to compute the integrals in question.

First, notice that the first integrand is even:

`(\cos(-x))/((-x)^2+1)=(\cos(x))/(x^2+1)`

`\implies\displaystyle\int_0^\infty(\cos(x))/(x^2+1)\,dx=\frac12\int_(-\infty)^\infty(\cos(x))/(x^2+1)\,dx`

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
`z=Re^(it)` with 0 ≤ t ≤ π.

Consider the complex function

`f(z)=(e^(iz))/(z^2+1)`

and notice that our original integrand is the real part of f(z). Then the integral of f(z) over C is

`\displaystyle\int_Cf(z)\,dz=\lim_(R\to\infty)\left(\int_(-R)^Rf(z)\,dz+\int_\Gamma f(z)\,dz\right)`

As R → ∞, the first integral on the right is exactly twice the one we want. Estimate the second one to be bounded by

`\displaystyle\left|\int_\Gamma f(z)\,dz\right|\le\pi R|f(z)|\le(\pi R)/(R^2-1)`

since

`|z^2+1|\ge\bigg||z^2|-|-1|\bigg|=|R^2-1|`

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,

`\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`

`\implies\displaystyle\int_0^\infty(\cos(x))/(x^2+1)\,dx=\frac12\mathrm{Re}\left(\int_Cf(z)\,dz\right)=\frac\pi{2e}`

For the second integral, we recall that for complex z,

`\sqrt z=\exp\left(\frac12\left(\ln|z|+i\arg(z)\right)\right)`

Consider a keyhole contour C, the union of

•
`\Gamma_R`, the larger circle with radius R and
`z=Re^(it)`, with 0 < t < 2π ;

•
`\Gamma_\varepsilon`, the smaller circle with radius ε and
`z=\varepsilon e^(-it)`, with 0 < t < 2π ;

•
`\ell_1`, the line segment above the positive real axis joining
`\Gamma_\varepsilon` to
`\Gamma_R` ; and

•
`\ell_2`, the other line segment below the positive real axis joining
`\Gamma_R` to
`\Gamma_\varepsilon`

Then

`\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`

and in the limit, the integral over
`\ell_1` converges to the one we want.

Estimate the integrals over the circular arcs:

•
`\Gamma_R` :

`\displaystyle\left|\int_(\Gamma_R)f(z)\,dz\right|\le2\pi R|f(Re^(it))|\le(2\pi R^(3/2))/(|R-\sqrt5|^2)\to0`

as R → ∞.

•
`\Gamma_\varepsilon` :

`\displaystyle\left|\int_(\Gamma_\varepsilon)f(z)\,dz\right|\le2\pi \varepsilon|f(\varepsilon e^(-it))|\le(2\pi\varepsilon^(3/2))/(|\varepsilon-\sqrt5|^2)\to0`

as ε → 0.

Consider the integral over
`\ell_2` :

`\displaystyle\int_(\ell_2)f(z)\,dz=\int_R^\varepsilon(\sqrt z)/(z^2+2z+5)\,dz\\\\=\int_R^\varepsilon(\exp\left(\frac12\left(\ln|z|+2\pi i\right)\right))/(z^2+2z+5)\,dz\\\\=-\int_R^\varepsilon(\exp\left(\frac12\ln|z|\right))/(z^2+2z+5)\,dz\\\\=\int_\varepsilon^R(\sqrt z)/(z^2+2z+5)\,dz\\\\=\int_(\ell_1)f(z)\,dz`

so in fact,

`\displaystyle\int_Cf(z)\,dz=2\int_0^\infty(\sqrt x)/(x^2+2x+5)\,dx`

By the residue theorem,

`\displaystyle\int_Cf(z)\,dz=2\pi i\sum_(\rm poles)\mathrm{Res}\,f(z)`

We have poles at z = -1 + 2i and z = -1 - 2i. On our chosen branch,

`√(-1+2i)=i\sqrt[4]{5}\exp\left(-\frac i2\tan^(-1)(2)\right)`

`√(-1-2i)=i\sqrt[4]{5}\exp\left(\frac i2\tan^(-1)(2)\right)`

The residues are

`\mathrm{Res}(f(z),z=-1-2i)=\frac{i\sqrt[4]{5}\exp\left(\frac i2\tan^(-1)(2)\right)}{-4i}`

`\mathrm{Res}(f(z),z=-1+2i)=\frac{i\sqrt[4]{5}\exp\left(-\frac i2\tan^(-1)(2)\right)}{4i}`

Their sum is

`\displaystyle\sum_(\rm poles)\mathrm{Res}\,f(z)=-\frac{\sqrt[4]{5}}2\sin\left(\frac12\tan^(-1)(2)\right)=-\frac{\sqrt[4]{5}}2\sqrt{(5-\sqrt5)/(10)}=-\frac i2√(\frac1\phi)`

where ɸ = (√5 + 1)/2 is the golden ratio, and so the overall integral is

`\displaystyle\int_0^\infty(\sqrt x)/(x^2+2x+5)\,dx=\frac\pi2√(\frac1\phi)`

Lastly, recall

`\displaystyle\sum_(k=0)^\infty\frac1{k!}=e`

Then our expression reduces to

`\left(\frac{e*\frac\pi{2e}}{\frac\pi2√(\frac1\phi)}\right)^2=\boxed{\phi}`