Integral $\int_0^{\infty}\cos(a_0+a_1x+a_2x^2)\frac{1}{x^2+\frac{1}{4}}dx$

Is this integral known to have a closed form?

$$\int_0^{\infty}\cos(a_0+a_1x+a_2x^2)\frac{1}{x^2+\frac{1}{4}}dx$$

Is there anything special about it?

For $\color{blue}{a_{_1}=0}$ and $\big\{a_{_0},a_{_2}\big\}\subset\mathbb R$, the integrand can always be rewritten as $~\dfrac{\cos\Big(Ax^2\pm B\Big)}{x^2+C},~$ where $A,B,C>0.~$ In this particular case, $C=1/4$.

\begin{align}I_+=\dfrac\pi{2\sqrt C}\bigg[\cos\big(B-AC\big)+F_{_S}\bigg(\sqrt{\dfrac{2ac}\pi}~\bigg)\cdot\Big(\sin\big(B-AC\big)-\cos\big(B-AC\big)\Big)\quad\\\\-F_{_C}\bigg(\sqrt{\dfrac{2ac}\pi}~\bigg)\cdot\Big(\sin\big(B-AC\big)+\cos\big(B-AC\big)\Big)\bigg].\end{align}

\begin{align}I_-=\dfrac\pi{2\sqrt C}\bigg[\cos\big(B+AC\big)+F_{_C}\bigg(\sqrt{\dfrac{2ac}\pi}~\bigg)\cdot\Big(\sin\big(B+AC\big)-\cos\big(B+AC\big)\Big)\quad\\\\-F_{_S}\bigg(\sqrt{\dfrac{2ac}\pi}~\bigg)\cdot\Big(\sin\big(B+AC\big)+\cos\big(B+AC\big)\Big)\bigg].\end{align}

Here, $F_{_S}$ and $F_{_C}$ represent the Fresnel sine and cosine integrals.

I doubt that there is a closed form in general. If $a_1 =0$, Maple finds a closed form involving the Anger J function:

$$1/4\,{\frac {\sqrt {2}\sqrt {\pi } \left( -\sin \left( a_{{0}} \right) {a_{{2}}}^{3/2}\pi -4\,\cos \left( a_{{0}} \right) \sqrt {a_{ {2}}}\pi \right) {{\rm \bf J}_{1/2}\left(1/4\,a_{{2}}\right)}}{a_{{2} }}}+1/4\,\sqrt {2}{\pi }^{3/2}\cos \left( a_{{0}} \right) \sqrt {a_{{2 }}}{{\rm \bf J}_{3/2}\left(1/4\,a_{{2}}\right)}+1/4\,{\frac {\sqrt {2} \sqrt {\pi } \left( 2\,\sqrt {2}\sin \left( 1/4\,a_{{2}} \right) \sin \left( a_{{0}} \right) \sqrt {\pi }a_{{2}}+2\,\sqrt {2}\cos \left( a_ {{0}} \right) \cos \left( 1/4\,a_{{2}} \right) \sqrt {\pi }a_{{2}}+8\, \cos \left( a_{{0}} \right) \sqrt {a_{{2}}} \right) }{a_{{2}}}}$$

• I think I made a mistake though. I would need $a_0$ replaced by x. Would that give a closed form? Really sorry about this. – KAT Feb 1 '15 at 3:08