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.
Best Answer
You're allowed to do this anytime the series is uniformly convergent.
In uniform convergence, you tell me the $\epsilon$ that you want the entire partial sum to be within for the entire graph, and I give you an $M$ that guarantees you can get within $\epsilon$ if you choose $n > M$, but the $M$ must work anywhere on the graph, and not be dependent on which point you choose. In regular (called pointwise) convergence, the $\epsilon-M$ guarantee can use a different $M$ at different points on the graph.
Pointwise vs. Uniform Convergence
https://proofwiki.org/wiki/Definition:Uniform_Convergence