Let $ F(x, a) = \int_0^x{\frac{t^p}{(t^2+a^2)^q}dt} $ Show that $ F(x, a) = a^{p+1-2q}F(x/a, 1) $ Using Apostol's Calculus and I'm trying to understand one of the questions:
Let $$ F(x, a) = \int_0^x{\frac{t^p}{(t^2+a^2)^q}dt} $$
where $a > 0$, and $p$ and $q$ are positive integers. Show that
$$ F(x, a) = a^{p+1-2q}F(x/a, 1) $$
What is $ F(x/a, 1) $? Do I just divide $x$ and $a$ in $ F(x, a) $ by $a$ so that we have $$ F(x/a, 1) = \int_0^{x/a}{\frac{t^p}{(t^2+1)^q}dt} $$
and if so, what next?
 A: As stated, the question doesn't make sense: One of the expressions depends on $q$, while the other does not.
What is true, however, is that
\begin{align*}
F\left(\frac x a, 1 \right) &= \int_0^{x/a} \frac{t^p}{t^2 + 1^2} \, dt \\
&= a^{1 - p} \int_0^{x/a} \frac{(at)^p}{(at)^2 + a^2} \,d(at) \\
&= a^{1 - p} \int_0^x \frac{t^p}{t^2 + a^2} \, dt \\
&= a^{1 - p} F(x, a)
\end{align*}

The answer above refers to an older version of the question, but the modification in the denominator is straightforward. Start by replacing
$$(t^2 + 1)^q = a^{-2q} \big((at)^2 + a^2\big)^q$$
and notice how the leading power of $a$ ought to change.
A: If $$F(x,a) = \int_{t=0}^x \frac{t^p}{(t^2+a^2)^q} \, dt,$$ then $$F(x/a,1) = \int_{t=0}^{x/a} \frac{t^p}{(t^2 + 1)^q} \, dt.$$  This suggests that the substitution $t = au$, $dt = a \, du$ changes $F(x,a)$ into $$\int_{u = 0}^{x/a} \frac{(au)^p}{((au)^2 + a^2)^q} \cdot a \, du = a^{p+1-2q} \int_{u=0}^{x/a} \frac{u^p}{(u^2+1)^q} \, du,$$ which is what you want to show.
