Proving $x\int_0^{\frac{a}{x}}f(xt)dt=\int_0^a f(t)dt$

Question

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!

Answer 1

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.

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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.

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This method works, but it's much longer than it needs to be.

Answer 2

Try to use the change of variable $s=xt$.