I have an engineering background. While reading signal processing, I came across this integral: $$\frac{1}{4}\int_{-\infty}^{\infty} X(u) Y^*(u) \text{d}u+\frac{1}{4}\int_{-\infty}^{\infty} X^*(v) Y(v)\text{d}v$$ $$=\frac{1}{2}Re\{\int_{-\infty}^{\infty} X(f) Y^*(f) \text{d}f\}$$
where $*$ implies complex conjugate and $Re$ means real part. How is this possible? I mean I am unable to find the missing steps, because the book does not gives the relation between $X(f)$ and $X(u)$ and $X(v)$, where $u$ and $v$ are dummy variables.
PS: $X(\cdot)$ is a low pass signal
There is nothing significant going on here. It is just simple algebra, and the fact that integration is linear with respect to the integrand.
If $x, y$ are any two complex numbers, then the conjugate of $xy^*$ is $x^*y$, and if you add a complex number and its conjugate, you get twice its real part: $(a+bi) + (a - bi) = 2a$.
So $$X(u)Y^*(u) + X^*(u)Y(u) = 2\,\text{Re}\bigg(X(u)Y^*(u)\bigg)\\ \frac14X(u)Y^*(u) + \frac14X^*(u)Y(u) = \frac12\,\text{Re}\bigg(X(u)Y^*(u)\bigg)\\ \frac14\int_{-\infty}^\infty X(u)Y^*(u)\,du + \frac14\int_{-\infty}^\infty X^*(u)Y(u)\,du = \frac12\,\text{Re}\left(\int_{-\infty}^\infty X(u)Y^*(u)\,du\right)$$
The final thing they did is to change the dummy variables in the three integrals to be different. Why, I have no idea, as dummy variables are just notational conveniences. The variables are only defined within the scope of their integral, and changing them does not affect the integral at all: $$\int_{-\infty}^\infty X^*(u)Y(u)\,du = \int_{-\infty}^\infty X^*(v)Y(v)\,dv\\\int_{-\infty}^\infty X(u)Y^*(u)\,du = \int_{-\infty}^\infty X(f)Y^*(f)\,df$$