What does e have to do with this integral?

This Integral can be solved by complex methods with a shorter way than my solution, but I like to solve it using elementary methods:

$\mathbf{Preposition \; 1 \; : }$

$\displaystyle \int_0^{\infty} \frac{1}{\sqrt{y}} e^{-(\frac{a}{y} + by ) } dy = \sqrt{\frac{\pi}{b}} e^{-2\sqrt{ab}}$

$\mathbf{Proof \; : }$

$\displaystyle A= \int_0^{\infty} \frac{1}{\sqrt{y}} e^{-(\frac{a}{y} + by ) } dy$

$\displaystyle \overset{{\color{#D61F06}{y \to \sqrt{\frac{a}{b}} y }} }{=} \sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \frac{1}{\sqrt{y}} e^{-\sqrt{ab} ( \frac{1}{y}+y)} dy$

$\displaystyle \overset{ {\color{#D61F06}{y \to \frac{1}{y} }} }{=}\sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \frac{1}{y\sqrt{y}} e^{-\sqrt{ab} ( \frac{1}{y}+y)} dy$

Hence:

$\displaystyle 2A = \sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \bigg( \frac{1}{\sqrt{y}} + \frac{1}{y\sqrt{y}} \bigg) e^{-\sqrt{ab} ( \frac{1}{y}+y)} dy$

$\displaystyle A = \sqrt{\sqrt{\frac{a}{b}}} \int_0^{\infty} \bigg( \frac{1}{2\sqrt{y}} + \frac{1}{2y\sqrt{y}} \bigg) e^{-\sqrt{ab} ( \sqrt{y} - \frac{1}{\sqrt{y}})^2 -2\sqrt{ab} } dy$

$\displaystyle \overset{{\color{#D61F06}{\sqrt{y} - \frac{1}{\sqrt{y}} \to y }}}{=} \sqrt{\sqrt{\frac{a}{b}}} \int_{-\infty}^{\infty} e^{-\sqrt{ab}y^2 - 2\sqrt{ab}} dy$

This can be derived from the Gaussian Integral :

$\displaystyle \boxed{A = \sqrt{\frac{\pi}{b}} e^{-2\sqrt{ab}} }$ .

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$\mathbf{Preposition \; 2 \; : }$

$\displaystyle \int_{-\infty}^{\infty} \frac{\cos (mx)}{x^2+1} dx = \pi e^{-m}$

$\mathbf{Proof \; : }$

Note that :

$\displaystyle \frac{1}{x^2+1} = \int_0^{\infty} e^{-(x^2+1)y} dy$

So:

$\displaystyle \int_{-\infty}^{\infty} \frac{\cos (mx)}{x^2+1} dx$

$\displaystyle = \int_0^{\infty} \int_{-\infty}^{\infty} \cos (mx) e^{-(x^2+1)y} dx dy$

$\displaystyle = \Re \int_0^{\infty} \int_{-\infty}^{\infty} e^{mix} e^{-(x^2+1)y} dx dy$

$\displaystyle = \Re \int_0^{\infty} \int_{-\infty}^{\infty} e^{-yx^2 +mix -y} dx dy$

Now use the formula:

$\displaystyle \int_{-\infty}^{\infty} e^{-ax^2+bx+c} dx = \sqrt{\frac{\pi}{a} } e^{\frac{b^2}{4a} +c}$

Which can be easily proved by the Gaussian Integral . To obtain:

$\displaystyle = \sqrt{\pi} \int_0^{\infty} \frac{1}{\sqrt{y}} e^{-(\frac{m^2}{4y} + y)} dy$

Now use Preposition 1 with $a= \frac{m^2}{4} \;\; \text{and} \;\; b= 1$ :

$\displaystyle\boxed{ = \pi e^{-m}}$

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$$

Now rewrite our integral as :

$\displaystyle \int_{-\infty}^{\infty} \frac{\cos^3 (x)}{x^2+1} dx$

$\displaystyle = \int_{-\infty}^{\infty} \frac{\frac{1}{4} \cos (3x)+ \frac{3}{4} \cos (x)}{x^2+1} dx$

$\displaystyle = \frac{1}{4} \int_{-\infty}^{\infty} \frac{\cos (3x)}{x^2+1} dx + \frac{3}{4} \int_{-\infty}^{\infty} \frac{\cos (x)}{x^2+1} dx$

$\displaystyle \overset{{\color{#D61F06}{\text{(Use Preposition 2)}}}}{=} \frac{1}{4} \pi e^{-3} + \frac{3}{4} \pi e^{-1}$

$\displaystyle \boxed{ = \frac{\pi^1}{4} ( \frac{1}{e^3} + \frac{3}{e^1} ) }$

Lemma :

$\displaystyle \int_{0}^{\infty} \frac{\cos mx}{x^2+a^2} \mathrm{d} x=\frac{\pi}{2a} e^{-am}$

Proof :

\text{J}(m) = \displaystyle \int_{0}^{\infty} \dfrac{\cos mx}{x^2+a^2} \mathrm{d} x \tag{1}

Using Integration by parts,

$\text{J}(m) = \displaystyle \int_{0}^{\infty} \dfrac{2x\sin mx}{m(x^2+a^2)^2} \mathrm{d} x$

\implies m \cdot \text{J}(m) = \displaystyle \int_{0}^{\infty} \dfrac{2x \sin mx}{(x^2+a^2)^2} \mathrm{d} x \tag {2}

Partially Differentiating $(2)$ w.r.t. $m$ , we have,

$\text{J} + m\dfrac{\partial}{\partial m} \text{J} = \displaystyle \int_{0}^{\infty} \dfrac {2x^2\cos(mx)}{(x^2+a^2)^2} \mathrm{d}x = \displaystyle \int_{0}^{\infty} \dfrac {2\cos(mx)}{(x^2+a^2)} \mathrm{d}x - 2a^2\displaystyle \int_{0}^{\infty}\dfrac{\cos mx}{(x^2+a^2)^2} \mathrm{d}x$

\implies a\dfrac{\partial}{\partial m} \text{J} = \text{J} - 2b^2\displaystyle \int_{0}^{\infty}\dfrac{\cos mx}{(x^2+a^2)^2} \mathrm{d}x \tag{3}

Again, partially differentiating $(3)$ w.r.t. $m$ , we have,

$m\dfrac{\partial^2}{\partial m^2} \text{J} = a^2 \displaystyle \int_{0}^{\infty} \dfrac{2x \sin mx}{(x^2+a^2)^2} \mathrm{d} x$

\implies \dfrac{\partial^2}{\partial m^2} \text{J} = a^2 \text{J} \tag{*} (from $(2)$ )

Note that $\text{J}(0) = \dfrac{\pi}{2a}$ and $\text{J}(\infty) = 0$

Now, solving $(*)$ , we have,

$\text{J}(m) = \dfrac{\pi}{2a} e^{-am} \quad \square$

Now,

$\displaystyle \text{I} = \int_{-\infty}^{\infty} \dfrac{\cos^3 (x)}{x^2+1} \mathrm{d}x = 2 \int_{0}^{\infty} \dfrac{\cos^3 (x)}{x^2+1} \mathrm{d}x$

$\displaystyle = \dfrac{1}{2} \int_{0}^{\infty} \frac{\cos (3x)}{x^2+1} \mathrm{d}x + \dfrac{3}{2} \int_{0}^{\infty} \dfrac{\cos (x)}{x^2+1} \mathrm{d}x$

Using the Lemma, we have,

$\displaystyle \text{I} = \dfrac{\pi}{4e^3} + \dfrac{3 \pi}{4e}$

$\implies p+q+r+s+t+u = \boxed{13}$