# Computing $f(x)$ if $\int_1^{xy} f(t) dt = y \int_1^x f(t) dt + x\int_1^y f(t) dt$ and $f(1) = 3$

This is a problem from Apostol, Calculus, Volume I, Chapter 6.9 (p. 238), that I was hoping someone could help with:

A function $f$, continuous on the positive real axis, has the property that $$\int_1^{xy} f(t) dt = y \int_1^x f(t) dt + x \int_1^y f(t) dt$$ for all $x > 0$ and all $y > 0$. If $f(1) = 3$, compute $f(x)$ for each $x > 0$.

My initial thought was to use properties of the integral to write:

\begin{align*} \int_1^{xy} f(t) dt & = \int_1^x f(t) dt + \int_x^{xy} f(t) dt \\ & = \int_1^x f(t) dt + x \int_1^y f(xt) dt \end{align*}

and

\begin{align*} \int_1^{xy} f(t) dt & = \int_1^y f(t) dt + \int_y^{xy} f(t) dt \\ & = \int_1^y f(t) dt + y \int_1^x f(yt) dt \end{align*}

Then each of these is also $=y \int_1^x f(t)dt + x \int_1^ f(t)dt$ from the given equation of the problem. I cannot seem to pull an equation for $f(x)$ out of this though (I might be missing some very simple manipulation...).

I also thought in terms of letting $A(x) = \int_1^x f(t)dt$, so the problem is giving us a functional equation: $$A(xy) = yA(x) + xA(y)$$

This has obvious similarities to the functional equation of the logarithm.

This problem is in the section of the text with the definition of the logarithm as an integral. I'm self-studying, so this isn't homework. Hints or full solutions are equally welcome. Thanks for the help.

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Fix $x$ and differentiate both sides of your equation with respect to $y$ to obtain: $$\tag{1} x f(xy)= \int_1^xf(t)\,dt +xf(y).$$ Set $y=1$ in equation $(1)$ to obtain $$\tag{2} xf(x)=3x+\int_1^x f(t)\,dt.$$ From $(2)$, we see that $f$ is differentiable. Unfixing $x$ now, differentiate both sides of $(2)$ with respect to $x$ to obtain $$f(x)+xf'(x)= 3+f(x).$$ So $$\tag{3} xf'(x)=3; \ \ f(1) =3.$$ The solution of the initial value problem $(3)$ is $f(x)= 3\ln x +3$.

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Ah! That's it. I knew there was something fairly straightforward that I was missing. Thanks. – user23784 Mar 24 '12 at 21:43
@David This is a great approach. – Pedro Tamaroff Mar 25 '12 at 2:00

$$A(xy) = yA(x) + xA(y)$$

Can be rewritten as

$$\frac{A(xy)}{xy}=\frac{A(x)}{x}+\frac{A(y)}{y} \,.$$

Let $B(x)=\frac{A(x)}{x}$ .

Then $B(xy)=B(x)+B(y)$. Which means that

$$B(e^{t+s})=B(e^{t}e^{s})= B(e^{t})+B(e^{s})\,.$$

Setting $C(t)=B(e^{t})$ you get that $C$ satisfies the Cauchy Equation

$$C(t+s)=C(t)+C(s) \,.$$

P.S. The equation of $B(x)$ is exactly the equation of the logarithm, you can actually stop there and figure which log it is...

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Sorry, I'm getting lost in here. We have not yet defined $e$ as a function, so I don't understand how you are making the step after $B(xy) = B(x) + B(y)$. I also don't know what is the Cauchy equation. Am I able to say $B(xy) = B(x) + B(y) \implies B$ is the logarithm? It isn't clear to me that some other function couldn't satisfy such a functional equation. Sorry, I feel like this should be obvious to me once you've rewritten in terms of $B(xy) = B(x) + B(y)$, but I'm just not seeing how to get an equation for $f(x)$ out of this. – user23784 Mar 24 '12 at 21:29
Well it depends how the logarithm is defined. It can be proven that if $B:(0, \infty) \rightarrow R$ is continuous and it satisfies $B(xy)=B(x)+B(y)$ then $B(x)= \log_a(x)$ for some $a$. This is easy to prove if you know the exponential functions and the Cauchy equation. – N. S. Mar 24 '12 at 21:35
The Cauchy equation is actually an easy exercise: Let $C: R \rightarrow R$ be a function so that $f(x+y)=f(x)+f(y)$. Let $f(1)=a$. Then you can prove that $f(n)=an$ for all integers $n$, and further that $f(\frac{p}{q})=a \frac{p}{q}$ for $p,q$ integer. It follows in particular that if $f$ is continuous, then $f(x)=ax$. – N. S. Mar 24 '12 at 21:38
Oh, that's pretty cool. I'll try to prove this. Seems like a good exercise. Though somehow I think I've drifted from the intent of the problem. Thanks though. Definitely interesting stuff to work on. – user23784 Mar 24 '12 at 21:42