What is the meaning of positive and negative shift in signal correlation? If $x(t)$ and $y(t)$ are real valued signal then from this page cross correlation between them is defined as

\begin{align}
R_{XY}(\tau) &amp;= \int_{-\infty}^{\infty}x(t)y(t-\tau) dt\quad\text{+ve shift}\tag{1a}\\
R_{XY}(\tau) &amp;= \int_{-\infty}^{\infty}x(t+\tau)y(t) dt\quad\text{-ve shift}\tag{1b}\\
\end{align}



I tried to solve a problem where $x(t) = e^{-at}u(t)$ and $y(t) = e^{-bt}u(t)$.


  From $(1a)$: Positive shift


\begin{align}
R_{XY}(\tau) &amp;= \int_{-\infty}^{\infty}x(t)y(t-\tau)dt\\
&amp;= \int_{-\infty}^{\infty}e^{-at}e^{-b(t-\tau)}u(t)u(t-\tau)dt\tag{2}
\end{align}
For $\tau\geq0$. Limits of integration will be from $\tau\leq t\leq\infty$. Hence from $(2)$ after simplification integration can be written as
\begin{align}
R_{XY}(\tau) &amp;= \int_{\tau}^{\infty}e^{-at}e^{-b(t-\tau)}dt\\
&amp;=e^{b\tau}\int_{\tau}^{\infty}e^{-(a+b)t}dt\\
&amp;=\frac{e^{-a\tau}}{a+b}
\end{align}
For $\tau\leq0$. Limits of integration will be from $0\leq t\leq\infty$. Hence from $(2)$ after simplification integration can be written as
\begin{align}
R_{XY}(\tau) &amp;= \int_{0}^{\infty}e^{-at}e^{-b(t-\tau)}dt\\
&amp;=-\frac{e^{b\tau}}{a+b}\int_{0}^{\infty}e^{-(a+b)t}dt\\
&amp;=\frac{e^{b\tau}}{a+b}
\end{align}
Therefore
\begin{align}
R_{XY}(\tau)=\begin{cases}
                    \frac{e^{-a\tau}}{a+b}; \tau\geq0\\
                    \frac{e^{b\tau}}{a+b}; \tau\leq0
             \end{cases}\tag{3}
\end{align}




  From $(1b)$: Negative shift


\begin{align}
R_{XY}(\tau) &amp;= \int_{-\infty}^{\infty}x(t+\tau)y(t)dt\\
&amp;= \int_{-\infty}^{\infty}e^{-a(t+\tau)}e^{-bt}u(t+\tau)u(t)dt\tag{4}
\end{align}
For $\tau\geq0$. Limits of integration will be from $0\leq t\leq\infty$. Hence from $(4)$ after simplification integration can be written as
\begin{align}
R_{XY}(\tau) &amp;= \int_{0}^{\infty}e^{-a(t+\tau)}e^{-bt)}dt\\
&amp;=e^{-a\tau}\int_{0}^{\infty}e^{-(a+b)t}dt\\
&amp;=\frac{e^{-a\tau}}{a+b}
\end{align}

Corrected my analysis after @Fat32 answer

For $\tau\leq0$. Limits of integration will be from $-\tau\leq t\leq\infty$. Hence from $(4)$ after simplification integration can be written as
\begin{align}
R_{XY}(\tau) &amp;= \int_{-\tau}^{\infty}e^{-a(t+\tau)}e^{-bt)}dt\\
&amp;=e^{-a\tau}\int_{-\tau}^{\infty}e^{-(a+b)t}dt\\
&amp;=\frac{e^{b\tau}}{a+b}
\end{align}
Therefore
\begin{align}
R_{XY}(\tau)=\begin{cases}
                    \frac{e^{-a\tau}}{a+b}; \tau\geq0\\
                    \frac{e^{b\tau}}{a+b}; \tau\leq0
             \end{cases}\tag{5}
\end{align}

Hence from $(3)$ and $(5)$ as can be seen the result match. If my analysis is wrong do correct me and if its correct then what does these negative and postive shift signifies?

On substituting $u = t - \tau \Rightarrow t = u + \tau \Rightarrow dt = du$ in $(1a)$, we get $(1b)$ as
\begin{align}
R_{XY}(\tau) &amp;= \int_{-\infty}^{\infty}x(u+\tau)y(u) du = \int_{-\infty}^{\infty}x(t+\tau)y(t) dt\\
\end{align}

Hence

\begin{align}
R_{XY}(\tau) &amp;= \int_{-\infty}^{\infty}x(t)y(t-\tau) du = \int_{-\infty}^{\infty}x(t+\tau)y(t) dt\\
\end{align}
Your Eq(3), for positive shift, is correct. And your Eq(5), for negative shift, should also be the same, but is wrong for $\tau &lt; 0$ case. 

The lower limit of the integral in Eq(4) should be $-\tau$ but you took it $\tau$. The reason is the step function $u(t+\tau)$ will have the nonzero range for $t &gt; -\tau$.

You will get the same result with the correct limit.