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I have an integral, and it looks simple enough for me to believe it could have an analytic solution; however I am unable to find it. I was trying "Gradshteyn and Ryzhik's Table of Integrals, Series, and Products", but it didn't help me either.

Maybe somebody of you can give me a hint:

$$f_{\pm}(x)=\int_0^{\theta_0} e^{-\frac{2 \theta^2}{\sigma^2}} \cdot \theta \cdot \Big( 1 + \cos\big(A\cdot[ x\pm B\cdot\theta]^2\big)\Big) d\theta$$

where $A,B,\sigma \in \mathbb{R}_+$ are constants, $\theta_0>\sigma$ (for instance, $\theta_0=4\sigma$) and $x \in (\mathbb{R}_+ \cup 0)$ is the argument of the function I want to find.


EDIT: I missed a square inside the cosine, sorry!

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  • $\begingroup$ See this link. $\endgroup$ – Intelligenti pauca Jan 24 '16 at 21:30
  • $\begingroup$ @Aretino: sorry, i was not careful, there was a square missing inside the cosine. Now wolfram doesnt find a solution anymore :-/ $\endgroup$ – Mario Krenn Jan 24 '16 at 21:45
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The following decomposition $$\int_0^{\theta_0} e^{-\frac{2 \theta^2}{\sigma^2}} \cdot \theta \cdot \Big( 1 + \cos(A\cdot x\pm B\cdot\theta)\Big) d\theta = \int_0^{\theta_0} e^{-\frac{2 \theta^2}{\sigma^2}} \cdot \theta d\theta+ Re\int_0^{\theta_0} e^{-\frac{2 \theta^2}{\sigma^2}} e^{iA(x\pm B\theta)^2} \theta d\theta $$ could help ? Each of the integrands of the right hand side has an explicit formula.

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  • $\begingroup$ I am sorry, I missed a square inside the coside. With this additional square, i think your very nice trick does not work anymore. sorry :-/ $\endgroup$ – Mario Krenn Jan 24 '16 at 21:42
  • $\begingroup$ This changes nothing. It still works ! $\endgroup$ – A. PI Jan 24 '16 at 21:56
  • $\begingroup$ Oh you are right, the right term still has an explicit representation. Great! Thanks a lot! $\endgroup$ – Mario Krenn Jan 24 '16 at 22:03

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