Proving $x\int_0^{\frac{a}{x}}f(xt)dt=\int_0^a f(t)dt$ Through some personal exploration I came upon what seems to be the fact that
$$x\int_0^{\frac{a}{x}}f(xt)dt=\int_0^a f(t)dt,\quad \forall x \in \mathbb{R} \setminus \{0\}.$$
I think a good first would be to prove that
$$\frac{d}{dx}x\int_0^{\frac{a}{x}}f(xt)dt=0,$$
but what little practical knowledge I have when it comes does not lend me to how to do that. Even if I did know how to prove that it was constant I still wouldn't know how to prove that
$$x\int_0^{\frac{a}{x}}f(xt)dt=\int_0^a f(t)dt.$$
Any pointers would be appreciated or resources that would be informative also be nice!
 A: Starting with
$$
x\int_0^{\frac{a}{x}}f(xt)dt,
$$
use the substitution $u=xt$.  In this case, $du=xdt$.  Moreover, if you substitute your $t$-limits, you get: when $t=0$, $u=x\cdot 0=0$ and when $t=\frac{a}{x}$, $u=x\cdot\frac{a}{x}=a$.  Therefore, the integral above equals
$$
\int_0^af(u)du.
$$
Changing variables from $u$ to $t$ gives your result.

With your approach, to show that
$$
\frac{d}{dx}\left[x\int_0^{\frac{a}{x}}f(xt)dt\right]=0,
$$
you will need to use the chain rule and product rule.  In particular, the derivative simplifies to
$$
\int_0^{\frac{a}{x}}f(xt)dt+x\frac{d}{dx}\left[\int_0^{\frac{a}{x}}f(xt)dt\right],
$$
after the product rule.  Now, applying the fundamental theorem of calculus, this simplifies to
$$
\int_0^{\frac{a}{x}}f(xt)dt+xf\left(x\cdot\frac{a}{x}\right)\frac{d}{dt}\left[\frac{a}{x}\right]+x\int_0^{\frac{a}{x}}\frac{d}{dx}[f(xt)]dt,
$$
This simplifies to
$$
\int_0^{\frac{a}{x}}f(xt)dt-\frac{a}{x}f(a)+x\int_0^{\frac{a}{x}}f'(xt)tdt,
$$
We can deal with this last integral using an integration by parts with
\begin{align}
u&=t&dv&=f'(xt)dt\\
du&=dt&v&=\frac{f(xt)}{x}.
\end{align}
Performing the integration by parts, we get
$$
\int_0^{\frac{a}{x}}f(xt)dt-\frac{a}{x}f(a)+x\left[\left.t\left(\frac{f(xt)}{x}\right)\right|_0^{\frac{a}{x}}-\int_0^{\frac{a}{x}}\frac{f(xt)}{x}dt\right].
$$
Substituting for $t$ at this point gives
$$
\int_0^{\frac{a}{x}}f(xt)dt-\frac{a}{x}f(a)+x\left[\frac{af(a)}{x^2}-\frac{1}{x}\int_0^{\frac{a}{x}}f(xt)dt\right].
$$
Distributing the $x$ at this point results in $0$.  Now you know that your integral is constant with respect to $x$, so it has the same value for all $x$'s.  If we choose $x$ to be $1$, then we'll get the same value as for any other $x$.  However, the $x=1$ case is equal to the simpler integral $\int_0^af(t)dt$.  Therefore, the two integrals are equal.

This method works, but it's much longer than it needs to be.
A: Try to use the change of variable $s=xt$.
