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Picture is from Ross' Introduction to Probability Models 11th ed.

Maybe a simple question, but I'm just wondering why the second equality in (5.17) follows. I tried substituting $y = t-s$, but that gives an extra factor of $-1$.

goblinb
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A substitution $y = t-s$ results in $\int_0^t G(t-s) ds = -\int_t^0 G(y)dy$. Notice how the limits of integration swapped. When $s$ is $0$, $y$ would be $t$ and vice versa. To put them in order, as in $\int_0^t$, you'd need to change the sign again.

user58697
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    Additional comment to help OP: the intergal on the left is positive and if you get a minus sign on RHS you would be equating a positive nubver to a negative one. This should alert you to a possible error in the way you changed the variable. – Kavi Rama Murthy Jun 17 '19 at 05:30