Interchanging summation with integral

Question

Is there any way to interchange between the sum and the integral:
$$\sum _{n=1,3,5,..}^{\infty} \frac{ \left[\int_0^{\sqrt{n}}\cos(s^2)ds\right]^2}{ n^3}=\frac{\pi^2}{16}$$
my goal is to get a closed form integral not series, the integral in the bracket is the well known Fresnel cosine integral $C(s)$

Answer 1

Yes, the expression can be written as a (double) integral with the sum done.

Let

$$f = \sum _{n=1,3,5,..}^{\infty} \frac{ \left(\int_0^{\sqrt{n}}\cos(s^2)ds\right)^2}{ n^3}\tag{1}$$

Now in the integral $i=\int_0^{\sqrt{n}}\cos(s^2)\,ds$ substitute $s=u \sqrt{n}$ which gives $i = \sqrt{n} \int_0^1\cos(n u^2)\,du$.

Hence the numerator of the summand of $f$ can be written as a double integral

$$g = \left(\int_0^{\sqrt{n}}\cos(s^2)ds\right)^2 = n \int_0^1\cos(n u^2)\,du \int_0^1\cos(n v^2)\,dv\tag{2}$$

Now interchanging sum and integrals and have to calculate the sum

$$s_{i}(u,v)=\sum _{n=1,3,5,..}^{\infty}\frac{1}{n^2} cos(n u^2) cos(n v^2)\tag{3}$$

the result is

$$s_{i}(u,v)=\frac{1}{16} e^{-i u^2-i v^2} \left(e^{2 i v^2} \Phi \left(e^{-2 i \left(u^2-v^2\right)},2,\frac{1}{2}\right)\\+e^{2 i u^2} \Phi \left(e^{2 i \left(u^2-v^2\right)},2,\frac{1}{2}\right)+\Phi \left(e^{-2 i \left(u^2+v^2\right)},2,\frac{1}{2}\right)\\+e^{2 i u^2+2 i v^2} \Phi \left(e^{2 i \left(u^2+v^2\right)},2,\frac{1}{2}\right)\right)\tag{4}$$

Here

$$\Phi (z,s,a)= \sum _{k=0}^{\infty } \frac{z^k}{(a+k)^s}$$

is the Lerch transcedent.

Finally $f$ can be written as a double integral

$$f = \int_0^1 \,du \int_0^1\,dv \;s_{i}(u,v)\tag{5}$$

Remark: because of the symmetry in the arguments the double integrals can possibly be transformed into single integrals.