Integral to series: $\int_0^1e^{\sin(\log x)}\ dx=1+\sum\limits_{n=1}^\infty\left(\prod_{k=1}^n\frac1{4k^2+1}-\prod_{k=1}^n\frac1{4k^2-4k+2}\right)$

I was looking for an explicit evaluation of the integral, when I stumbled across this series, that seems to converge to the value of the integral.

$$\mathcal I=\int_0^1e^{\sin(\log x)}\ dx=1+\sum_{n=1}^\infty\left(\prod_{k=1}^n\frac1{4k^2+1}-\prod_{k=1}^n\frac1{4k^2-4k+2}\right)$$

Additionally, the convergence is quite fast: summing up to $n=10$ gives $20+$ correct decimal digits!

To prove it, I tried to use the fact that $e^x=\sum\limits_{n=1}^\infty\frac{x^n}{n!}$ to get$$\mathcal I=\sum\limits_{n=1}^\infty\frac{1}{n!}\underbrace{\int_0^1\sin^n(\log x)\ dx}_{J_n}$$Now, to evaluate $J_n$, I used the fact that $\sin(\log x)=\frac{x^i-x^{-i}}{2i}$:\begin{align}J_n=\int_0^1\sin^n(\log x)\ dx & =\int_0^1\left(\frac{x^i-x^{-i}}{2i} \right)^n\ dx \\& =\frac{(-1)^n}{2^n}i^n\int_0^1\left(x^i-x^{-i}\right)^n\ dx\\& =\frac{(-1)^n}{2^n}i^n\int_0^1\sum_{k=0}^n{n \choose k}(-1)^kx^{in-2ik}\ dx \\& =\frac{(-1)^n}{2^n}i^n\sum_{k=0}^n{n \choose k}\frac{(-1)^k}{1+i(n-2k)}\\\end{align}and now what? I don't see how to make the $1$ pop out neither the two products. I tried to use the fact that $\frac1{1+iM}=\frac1{1+M^2}-i\frac{M}{1+M^2}$ in order to cancel the imaginary part sum (since the integral is real), but the presence of the $i^n$ in the front makes it difficut, and requires splitting into even and odd terms. For example, I got:$$J_{2m}=\int_0^1\sin^{2m}(\log x)\ dx=\frac{(-1)^m}{4^m}\sum_{k=0}^{2m}{{2m}\choose k}\frac{(-1)^k}{1+4(m-k)^2}$$and$$J_{2m+1}=\int_0^1\sin^{2m+1}(\log x)\ dx=-\frac{(-1)^m}{2^{2m+1}}\sum_{k=0}^{2m+1}{{2m+1}\choose k}\frac{(-1)^k(2m-2k+1)}{1+(2m-2k+2)^2}$$which are purely real. However I strongly believe this is not the way to go, since I don't see how to transform these two ugly sums in the more elegant products that appear in the original sum.There must be another way that I am missing.

I also tried the substitution $t=-\log x$, that leads to$$\mathcal I=\int_0^\infty e^{-(t+\sin t)}\ dt$$but from here I don't know how to proceed.

Any ideas? I am looking for a proof from the integral to the series, not for a closed form.