So I recently thought of this random integral, as I was thinking of integrals that have nice cancellation properties involving the fractional part. I believe I have an answer but my method is rather non-rigorous as I do not prove convergence, nor even know if it converges. Here is what I did:
Let $$J=\int_0^{\infty}\{e^x\}-\frac {1}{2}dx$$ Expressing it in terms of the floor function and noticing that for $x\in(\log(n),\log(n+1))$, you have $\lfloor e^x \rfloor=n$ yields
$J=\sum_{n=1}^\infty\int_{\log(n)}^{\log(n+1)}(e^x-\lfloor e^x\rfloor-\frac {1}{2} ) dx= \sum_{n=1}^\infty\int_{\log(n)}^{\log(n+1)}(e^x-n-\frac {1}{2} ) dx \\= \sum_{n=1}^\infty (1+(n+\frac{1}{2})\log(\frac{n}{n+1}))$
This can be expressed as
$$J= \lim_{N\to\infty}\sum_{n=1}^N((n-\frac{1}{2})\log(n)-(n+\frac{1}{2})\log(n+1)+\log(n)+1)$$ The first two terms are telescoping which gives
$J=\lim_{N\to\infty}(\log(N!)+N-(N+\frac{1}{2})\log(N+1))$
$ =\lim_{N\to\infty}(N\log(N)+\frac{1}{2}\log(2\pi N)-(N+\frac{1}{2})\log(N+1))$ (using Stirling's formula)
$ = \lim_{N\to\infty}((N+\frac{1}{2})\log(N)-(N+\frac{1}{2})\log(N+1))+\frac{1}{2}\log(2 \pi)=\frac{1}{2}\log(2 \pi)+\log(\frac{1}{e})=\frac{1}{2}\log(2 \pi)-1.$
It would be much appreciated if anyone could rectify any mistakes I made, make this rigorous and/or correct me on why the integral diverges.
Here is a generalization to an increasing function instead of $e^x$. An expression is derived for $\int_0^{\infty}(\{f(x)\}-\frac12)dx$ which, when specialized to $f(x)=e^x$, shows that the iimit does exist.
The expression for $J(n) =\int_0^{f^{(-1)}(n))}(\{f(x)\}-\frac12)dx$ is $\int_{m}^{n}yg'(y)dy-(n+\frac12)x_n+\sum_{k=m+1}^{n}x_{k} +c$ where $f(x_k)=k$ and $g$ is the inverse function of $f$ and $c$ and $m$ are constants independent if $n$.
I would not be surprised if the expression for $J(n)$ could be simplified by using some properties of $f(x)$.
Here is the moderately messy derivation.
Consider $J =\int_0^{\infty}(\{f(x)\}-\frac12)dx =\lim_{t \to \infty}\int_0^t(\{f(x)\}-\frac12)dx $.
Suppose $f(x) \ge, 0, f'(x) > 0, f''(x) > 0, f(x)\to \infty $ as $x \to \infty$, and $f$ has no singularities.
What can be said about $J$?
Let $g$ be the inverse of $f$ so $f(g(x))=g(f(x))=x $.
Let $x_n$ be such that $f(x_n)=n$ for positive integral $n$ and let $m$ be the smallest such that $f(x_m) > 0$.
Consider
$\begin{array}\\ J(n) &=\int_{x_m}^{x_n}(\{f(x)\}-\frac12)dx\\ &=\sum_{k=m}^{n-1}\int_{x_k}^{x_{k+1}}(\{f(x)\}-\frac12)dx\\ &=\sum_{k=m}^{n-1}\int_{x_k}^{x_{k+1}}(f(x)-\lfloor f(x)\rfloor-\frac12)dx\\ &=\sum_{k=m}^{n-1}\int_{x_k}^{x_{k+1}}(f(x)-k-\frac12)dx\\ &=\sum_{k=m}^{n-1}\int_{x_k}^{x_{k+1}}f(x)dx-\sum_{k=m}^{n-1}\int_{x_k}^{x_{k+1}}(k+\frac12)dx\\ &=\int_{m}^{n}f(x)dx-\sum_{k=m}^{n-1}(x_{k+1}-x_k)(k+\frac12)\\ &=\int_{m}^{n}f(x)dx-\sum_{k=m}^{n-1}k(x_{k+1}-x_k)-\frac12\sum_{k=m}^{n-1}(x_{k+1}-x_k)\\ &=\int_{m}^{n}f(x)dx-\frac12(x_n-x_m)-\sum_{k=m}^{n-1}kx_{k+1}+\sum_{k=m}^{n-1}kx_k\\ &=\int_{m}^{n}f(x)dx-\frac12(x_n-x_m)-\sum_{k=m}^{n-1}(k+1-1)x_{k+1}+\sum_{k=m}^{n-1}kx_k\\ &=\int_{m}^{n}f(x)dx-\frac12(x_n-x_m)-\sum_{k=m}^{n-1}(k+1)x_{k+1}+\sum_{k=m}^{n-1}x_{k+1}+\sum_{k=m}^{n-1}kx_k\\ &=\int_{m}^{n}f(x)dx-\frac12(x_n-x_m)-\sum_{k=m+1}^{n}kx_{k}+\sum_{k=m}^{n-1}x_{k+1}+\sum_{k=m}^{n-1}kx_k\\ &=\int_{m}^{n}f(x)dx-\frac12(x_n-x_m)-nx_n+mx_m+\sum_{k=m}^{n-1}x_{k+1}\\ &=\int_{m}^{n}f(g(y))g'(y)dy-\frac12(x_n-x_m)-nx_n+mx_m+\sum_{k=m+1}^{n}x_{k} \qquad x=g(y), dx=g'(y) dy\\ &=\int_{m}^{n}yg'(y)dy-\frac12(x_n-x_m)-nx_n+mx_m+\sum_{k=m+1}^{n}x_{k}\\ &=\int_{m}^{n}yg'(y)dy-(n+\frac12)x_n+\sum_{k=m+1}^{n}x_{k}+(m+\frac12)x_m\\ \end{array} $
If $f(x)=e^x, g(y)=\ln(y), g'(y)=\dfrac1{y}, x_n=\ln(n) $ so $m=1, x_m=0$
$\begin{array}\\ J(n) &=\int_{m}^{n}yg'(y)dy-(n+\frac12)x_n+\sum_{k=m+1}^{n}x_{k}+(m+\frac12)x_m\\ &=\int_{m}^{n}dy-(n+\frac12)\ln(n)+\sum_{k=m+1}^{n}\ln(k)+(m+\frac12)\ln(m)\\ &=n-m-(n+\frac12)\ln(n)+\ln(n!)-\ln(m!)+(m+\frac12)\ln(m)\\ &=n-(n+\frac12)\ln(n)+(n+\frac12)\ln(n)-n+c+O(\frac1{n}) \qquad c=\frac12\ln(\pi)+\text{terms involving }m\\ &=c\\ \end{array} $
So $J(n)$ does approach a limit as $n \to \infty$.