Can you generalise it

This is how I did this.

First, consider $\displaystyle I'(a)=2a \int_{0}^{\infty} \cos(a^{2}x^{2})\sin(\dfrac{1}{x^{2}})dx$

Next, we let $\displaystyle x=\dfrac{u}{\sqrt{a}}$ (This is where we need $a$ to be positive and also where the symmetry $a->-a$ breaks.)

$\displaystyle I'(a)=2\sqrt{a} \int_{0}^{\infty} \cos(ax^{2})\sin(\dfrac{a}{x^{2}})dx$ (Note that we also change a dummy variable from $u$ to $x$ at this step.)

Now, the idea is the sine and the cosine will split into a sum of 2 functions, one is symmetric under $\displaystyle x->\dfrac{1}{x}$ and the other is antisymmetric, in a sense that $\displaystyle f(x)="-"f(\dfrac{1}{x})$ , so, we need something that is invariant under the change of variable $\displaystyle x->\dfrac{1}{x}$

Consider, $\displaystyle J(a)=\int_{0}^{\infty} \cos(a^{2}x^{2})\sin(\dfrac{1}{x^{2}})\dfrac{1}{x^{2}}dx$

If we let $\displaystyle x=\dfrac{u}{\sqrt{a}}$ , we get

$\displaystyle J(a)=\sqrt{a} \int_{0}^{\infty} \cos(ax^{2})\sin(\dfrac{a}{x^{2}})\dfrac{1}{x^{2}}dx$

$\displaystyle \dfrac{I'(a)}{2\sqrt{a}}+\dfrac{J(a)}{\sqrt{a}}=\dfrac{1}{2}\int_{0}^{\infty} \left(\sin(ax^{2}+\dfrac{a}{x^{2}})+\sin(ax^{2}-\dfrac{a}{x^{2}}) \right) \left(1+\dfrac{1}{x^{2}} \right)dx$

Now, the second term on the last equality is antisymmetric under $\displaystyle x->\dfrac{1}{x}$ and so, vanishes and $\displaystyle (1+\dfrac{1}{x^{2}})dx=d(1-\dfrac{1}{x})$ , so, the first term is $\displaystyle \sqrt{\dfrac{\pi}{2a}}(\sin(2a)+\cos(2a))$

Next, I claim that $\displaystyle J'(a)=-2I(a)$ and hence, we arrive at

$\displaystyle J"(a)-4J(a)=-\sqrt{2\pi}(\sin(2a)+\cos(2a))$ this equation is solved by the usual method and we get

$\displaystyle J(a)=A\sinh(2a)+B\cosh(2a)+\sqrt{\dfrac{\pi}{32}}(\sin(2a)+\cos(2a))$ and since $J'(a)/a->0$ as $a->0$ so, we get $J(a)=\sqrt{\dfrac{\pi}{32}}(e^{-2a}+(\sin(2a)+\cos(2a)))$ and since $I(a)=-\dfrac{1}{2} J'(a)=\sqrt{\dfrac{\pi}{32}}(e^{-2a}+\sin(2a)-\cos(2a)))$

So, $\displaystyle A=32,B=-2,C=D=2$ and $A+B+C+D=\boxed{34}$

Now, we need to prove that $\displaystyle J'(a)=-2I(a)$

$\displaystyle J'(a)=-2a\int_{0}^{\infty} \sin(a^{2}x^{2})\sin(\dfrac{1}{x^{2}})dx$

By changing $\displaystyle x->\dfrac{1}{au}$ , we get the required result.