When is true that $\int_0^1 \int_0^1 f(xy)\ dx\ dy=\ -\int_0^1 \log(x) f(x)\ dx$

Question

Saw in a post this identity $$\int_0^1 \int_0^1 f(xy)\ dx\ dy=\ -\int_0^1 \log(x) f(x)\ dx$$ then the author says "notice that $f$ is under some restrictions", and concludes the post, without saying what restrictions! I'd like to know what these restrictions are, since in the proof he provided I couldn't find any.

Here is the proof:

Let $xy=t$, so $dx=\frac{dt}{y}$, and the integral becomes: $$\int_0^1 \int_0^y f(t)\ \frac{dt}{y}\ dy$$= $$\int_0^1 \frac{1}{y} \int_0^y f(t)\ dt\ dy$$ now integrate by parts with $$u=\int_0^yf(t)\ dt\hspace{2cm} dv=\frac{1}{y}\ dy$$ and get $$\left[\log(y)\int_0^yf(t)\ dt\right]_{y=0}^{y=1} -\int_0^1 \log(y) f(y)\ dy=$$ $$=\ 0\ -\ 0\ -\int_0^1 \log(y) f(y)\ dy=$$ $$=-\int_0^1 \log(y) f(y)\ dy$$ and the proof is concluded. Please help with the restrictions on $f$.

Answer 1

Here is a more general result:

Theorem. Let $f : [0, 1] \to \mathbb{R}$ be a measurable function. Then

$$ \int_{0}^{1}\int_{0}^{1} f(xy) \, \mathrm{d}x\mathrm{d}y = \int_{0}^{1} f(t)\log(1/t) \, \mathrm{d}t. \tag{*} $$

holds whenever at least one of the integrals in $\text{(*)}$ is finite with $f$ replaced by either

• the positive part of $f$, namely $f_+(t) = \max\{0, f(t)\}$, or
• the negative part of $f$, namely $f_-(t) = \max\{0, -f(t)\}$.

Proof. Define the two Borel measures $\mu$ and $\nu$ on $[0, 1]$ by

$$ \mu(E) = \int_{0}^{1}\int_{0}^{1} \mathbf{1}_{E}(xy) \, \mathrm{d}x\mathrm{d}y \qquad\text{and}\qquad \nu(E) = \int_{0}^{1} \mathbf{1}_{E}(t)\log(1/t) \, \mathrm{d}t. $$

Then for $0 \leq a \leq 1$, a straightforward computation shows that

\begin{align*} \mu([a, 1]) &= \int_{a}^{1}\int_{a/y}^{1} \mathrm{d}x\mathrm{d}y = 1 - a - a\log a = \int_{a}^{1} \log(1/t) \, \mathrm{d}t = \nu([a, 1]). \end{align*}

Since the sets of the form $[a, 1]$, $0 \leq a \leq 1$, generates the Borel $\sigma$-algebra $\mathcal{B}([0, 1])$, it follows that $\mu = \nu$. Therefore the desired conclusion follows. $\square$

Answer 2

In addition to the standard restriction that $f$ be integrable on $[0,1]$, another potential restriction on $f$ occurs where you used that

$$\left[\log(y)\int_0^y f(t)\,dt\right]_{y=0}^{y=1} = 0 - 0$$

In particular, there's a possible issue with

$$\lim_{y\to 0^{+}}\left(\log(y)\int_0^y f(t)\,dt\right) = 0$$

If $f(t)$ satisfies standard Riemann integrability, then the function is bounded in the range, so the limit above always applies (e.g., as explained in RRL's answer). However, if $f(t)$ can be something like an improper Riemann integrable function, then the above limit may not hold (e.g., where $\int_0^y f(t)\,dt = (-\log(y)+1)^{-\frac{1}{2}}$).

Answer 3

If $f$ is Riemann integrable on $[0,1]$, then no other restriction on $f$ is required in order that

$$\tag{*}\lim_{y \to 0+}\log(y) \int_0^y f(t) \, dt = 0$$

Since a Riemann integrable function is bounded, there exists $M>0$ such that $|f(t)| \leqslant M$ for all $t\in [0,1]$ and, hence, for all $y \in [0,1)$

$$0 \leqslant \left|\log(y) \int_0^y f(t) \, dt \right|= -\log(y)\left|\int_0^y f(t) \, dt\right|\leqslant -\log(y)\int_0^y|f(t)| \, dt\leqslant -My\log(y)$$

Since $\lim_{y \to 0+} y \log(y) = 0$, it follows from the squeeze theorem that (*) holds.

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If anything warrants concern it is the step $du = f(y) \,dy$ when $u = \int_0^y f(t) \,dt$. This is because $f$ need not be continuous at every point in $[0,1]$ when it is assumed only to be Riemann integrable. However, $f$ will be continuous almost everywhere and the integration by parts formula is still valid.