Integral of Fresnel functions

[johnglenn]

I have been working on this for several days but getting nowhere. Any help would be great.

\begin{align}
&\int_0^x dy\,y^2 \cos(y^2) C^2 \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right)
\end{align}

In reality only the first one is causing me troubles, however I have pasted the entire expression as it might lead to some cancellations. Any help would be welcome. $C()$ is the Fresnel integral.

 

[jvicens]

Hello there,

I can see that you are struggling with an integral involving the Fresnel integral. This type of integral can be tricky, but there are some techniques that you can use to solve it.

Firstly, let's take a look at the integral you have provided:
\begin{align}
&\int_0^x dy\,y^2 \cos(y^2) C^2 \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right)
\end{align}

One approach you can take is to use integration by parts. This involves rewriting the integral as a product of two functions and then applying the integration by parts formula. In this case, we can choose $u = y^2 \cos(y^2)$ and $dv = C^2 \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right)dy$. This will give us:
\begin{align}
&\int_0^x dy\,y^2 \cos(y^2) C^2 \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right) = \left[y^2 C^2 \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right)\right]_0^x - \int_0^x dy\,2yC \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right)\sin(y^2)
\end{align}

Notice that the first term on the right-hand side can be easily evaluated, leaving us with the integral:
\begin{align}
&\int_0^x dy\,2yC \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right)\sin(y^2)
\end{align}

This integral can be solved using a substitution, where we let $u = y^2$ and $du = 2ydy$. This will give us:
\begin{align}
&\int_0^x dy\,2yC \!\!\left(\!\frac{\sqrt{2}\,y}{\sqrt\pi}\!\right)\sin(y^2) = \int_0^{x^2} du\,C \!\