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

Question

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.

Answer 1

Expanding into even/odd functions and applying Maclaurin series noting $(-1)^{2n}=1,n\in\Bbb N$:

$$\int_0^\infty e^{-(x+\sin(x))}dx=\int_0^\infty e^{-x}\cosh(x)dx-\int_0^\infty e^{-x}\sinh(x)dx=\sum_{n=0}^\infty(-1)^{2n} \int_0^\infty\frac{e^{-x}\sin(x)^{2n}(x)}{(2n)!}dx-\sum_{n=0}^\infty (-1)^{2n+1}\int_0^\infty\frac{e^{-x}\sin^{2n+1}(x)}{(2n+1)!}dx$$

and use @JJacquelin’s findings:

$$\int_0^\infty\sin^{2n}(x)e^{-x}dx=\frac{(2n)!}{(1+2^2)(1+4^2)(1+6^2)\dots(1+(2n)^2)}$$ $$\int_0^\infty\sin^{2n+1}(x)e^{-x}dx=\frac{(2n+1)!}{(1+1^2)(1+3^2)(1+5^2)\dots(1+(2n+1)^2)}=\frac{(2n+1)!}{(1+(2(1)-1)^2)\dots(1+(2(n+1)-1)^2)}$$ From H.Bateman, "Tables of Integral Transforms", Eq.(3) & (7),p .150 , McGraw Hill, 1954.

The $(2n)!,(2n+1)!$ cancel out too. We get:

$$\int_0^\infty e^{-(x+\sin(x))}dx=\sum_{n=0}^\infty\prod_{j=1}^n\frac1{(2j)^2+1}-\sum_{n=0}^\infty\prod_{j=1}^{n+1}\frac1{(2j-1)^2+1}= 1+\sum_{n=1}^\infty\prod_{j=1}^n\frac1{(2j)^2+1}-\sum_{n=1}^\infty\prod_{j=1}^n\frac1{(2j-1)^2+1} $$

after substituting $n+1\to n$ in the second term and extracting the first term’s $n=0$ value, which is $1$ due to the empty product. Therefore:

$$\boxed{\int_0^\infty e^{-(x+\sin(x))}dx= 1+\sum_{n=1}^\infty\left(\prod_{j=1}^n\frac1{(2j)^2+1}-\prod_{j=1}^n\frac1{(2j-1)^2+1}\right)}$$