Does a definite integral equal to the Möbius function exist?

Sure. Fix any $a<b$, and put $f(x,n)=\tfrac{1}{b-a}\mu(n)$ (independent of $x$).

As you stated it you can define $f(x,n)$ is many ways to generate any sequence.

A more interesting answer is that what we want is a generating function of the sequence $\mu(n)$, and that the natural ones are $$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}$$ The partial sums admitting explicit formulas in term of the non-trivial zeros $$f(s)=\sum_{n \le x} \mu(n) n^{-s}$$ The inverse Mellin transform of $\frac{\Gamma(s)}{\zeta(s)}$ (again admitting an explicit formula) $$\sum_{n=1}^\infty \mu(n) e^{-nx}$$ Its periodic versions $$F(x)=\sum_{n=1}^\infty \frac{\mu(n)}{n^s} e^{inx}$$ and the weird Riesz function $$\sum_{n=1}^\infty \frac{\mu(n)}{n^2} e^{-x/n^2} = \sum_{k=0}^\infty \frac{(-x)^k}{k!} \frac{1}{\zeta(2+2k)}=\sum_{k=0}^\infty \frac{(-x)^k}{k!} \frac{1}{ \frac{(-1)^{k+1}(2\pi)^{2k} B_{2k}}{(2k)!}}$$ All the other useful generating functions of $\mu(n)$ are more or less directly related to one of them.

If we allow bending the rules a bit to study the Mertens function $$M(x)=\sum_{n=1}^{\lfloor x \rfloor}\mu(x)$$ and improper integral representations, then there's the Perron formula $$M(x)=\lim_{\sigma\to 1^+}\int_{-\infty}^{\infty}\frac{\mathrm{d}t}{2\pi}\frac{x^{\sigma+\mathrm{i}t}}{(\sigma+\mathrm{i}t)\zeta(\sigma+\mathrm{i}t)}\text{.}$$