Thank you Fresnel!

$\begin{aligned} I & = \int_0^\infty \sin(x^2) \left(\frac {x+4}x\right)^2 dx \\ & = \int_0^\infty \sin(x^2) \left(1+\frac 4x\right)^2 dx \\ & = \int_0^\infty \sin(x^2) \left(1+ \frac 8x + \frac {16}{x^2} \right) dx \\ & = \int_0^\infty \sin(x^2) \ dx + 8 \int_0^\infty \frac {\sin(x^2)}x \ dx + 16 \int_0^\infty \frac {\sin(x^2)}{x^2} \ dx \\ & = I_1 + 8I_2 + 16I_3 \end{aligned}$

$\begin{aligned} I_1 & = \int_0^{\color{#3D99F6}\infty} \sin(x^2) \ dx = {\color{#3D99F6}S(\infty)} = \sqrt{\frac \pi 8} & \small \color{#3D99F6} S(x) \text{ is Fresnel integral.} \end{aligned}$

$\begin{aligned} I_2 & = \int_0^\infty \frac {\sin(x^2)}x \ dx & \small \color{#3D99F6} \text{Let }u = x^2 \implies du = 2x \ dx \\ & = \frac 12 \int_0^ {\color{#3D99F6}\infty} \frac {\sin u}u \ du \\ & = \frac 12 \color{#3D99F6}Si (\infty) = \frac \pi 4 & \small \color{#3D99F6} Si(x) \text{ is sine integral.} \end{aligned}$

$\begin{aligned} I_3 & = \int_0^\infty \frac {\sin(x^2)}{x^2} \ dx & \small \color{#3D99F6} \text{By integration by parts.} \\ & = -\frac {\sin(x^2)}x \ \bigg|_0^\infty + \int_0^\infty \frac {2x\cos (x^2)}x\ dx \\ & = -0+{\color{#3D99F6}0} + 2 \int_0^{\color{#D61F06}\infty} \cos (x^2) \ dx & \small \color{#3D99F6} \text{See note: }\lim_{x \to 0} \frac {\sin (x^2)}x = 0 \\ & = 2 {\color{#3D99F6}S(\infty)} = \sqrt{\frac \pi 2} & \small \color{#3D99F6} C(x) \text{ is Fresnel integral.} \end{aligned}$

Therefore, $\displaystyle I = I_1 + 8I_2 + 16I_3 = \sqrt{\frac \pi 8} + 8 \times \frac \pi 4 + 16 \sqrt{\frac \pi 2} = 2\pi + \sqrt{\frac {1089}8 \pi}$

$\implies a+b+c = 2+1089+8 = \boxed{1099}$

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Note:

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin (x^2)}x & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {2x \cos (x^2)}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x. \\ & = 0 \end{aligned}$

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References:

• Fresnel integrals
• Sine integral
• L'Hôpital's rule