Is this integral known to have a closed form?


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:

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

  • $\begingroup$ 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. $\endgroup$ – KAT Feb 1 '15 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.