Find maximum value of $\int_{1}^3\frac{f(x)}{x}dx$ if $\int_{1}^{3}f(x)dx=0$ and also $-1\leq f(x)\leq 1$ Let $f$ be function such that $$\int_{1}^{3}f(x)dx=0$$ and $$-1\leq f(x)\leq 1$$ Then find the maximum value of $$\int_{1}^{3}\frac{f(x)}{x}dx$$
My Attempt:
I wonder if this approach is correct or not.
Let $$f(x)=\begin{cases} -c, 1\le x\le a \\ 1, a\le x<3 \end{cases}$$
As per the constraint $$c=\frac{3-a}{a-1}$$
Let $$G(a)=\int_{1}^{3}\frac{f(x)}{x}dx=\int_{1}^{a}\frac{-c}{x}dx+\int_{a}^{3}\frac{1}{x}dx=\frac{2}{1-a}\ln a+\ln 3$$
$$G'(a)=\frac{1-a+a\ln a}{a(1-a)^2}$$
After putting $G'(a)=0$ 
I am not able to solve for $a$
 A: This was an exercise in the 2014 Putnam and the trick is to write for a parameter $a>0$,
\begin{split}
\left\lvert\int_1^3 \frac{f(x)}x\,\mathrm dx\right\rvert &= \left\lvert\int_1^3\frac{f(x)}x\,\mathrm dx -\int_1^3\frac{f(x)}a\,\mathrm dx\right\rvert\\
&=\left\lvert\int_1^3 f(x)\left(\frac1x-\frac1a\right)\,\mathrm dx\right\rvert \\
&\le\int_1^3 \lvert f(x)\rvert\left\lvert\frac1x-\frac1a\right\rvert\,\mathrm dx \\
&\le\int_1^3 \left\lvert\frac1x-\frac1a\right\rvert\,\mathrm dx \\
&=\ln\left(\frac{a^2}3\right)+\frac4a-2.
\end{split}
Minimizing the last expression with $a=2$ we get $$\left\lvert\int_1^3 \frac{f(x)}x\,\mathrm dx\right\rvert\le\ln(4)-\ln(3).$$
Equality is achieved for the function suggested by @Student in the comments: $f(x)=1$ on $[1,2]$ and $f(x)=-1$ on $[2,3]$.
A: As usual, 
I will generalize.
Let $f$ be function such that 
$\int_{a}^{b}f(x)dx=0$
and 
$|f(x)|\leq 1$.
Find the best bound of 
$|\int_{a}^{b}f(x)g(x)dx|$
where
$g(x) > 0$
and
$g'(x) < 0$.
The original problem is
$a=1, b=3, g(x) = \dfrac1{x}$.
For
$g(a) > c > g(b)$,
$\begin{array}\\
|\int_{a}^{b}f(x)g(x)dx|
&=|\int_{a}^{b}f(x)g(x)dx-c\int_{a}^{b}f(x)dx|\\
&=|\int_{a}^{b}f(x)(g(x)-c)dx|\\
&\le\int_{a}^{b}|g(x)-c|dx\\
&\le\int_{a}^{g^{-1}(c)}|g(x)-c|dx+\int_{g^{-1}(c)}^{b}|g(x)-c|dx\\
&=\int_{a}^{g^{-1}(c)}(g(x)-c)dx+\int_{g^{-1}(c)}^{b}(c-g(x))dx\\
&=c(b-2g^{-1}(c)+a)+\int_{a}^{g^{-1}(c)}g(x)dx-\int_{g^{-1}(c)}^{b}g(x)dx\\
&=c(b-2h(c)+a)+\int_{a}^{h(c)}g(x)dx-\int_{h(c)}^{b}g(x)dx
\qquad h(x) = g^{-1}(x)\\
&=c(b-2h(c)+a)+G(h(c))-G(a)-G(b)+G(h(c))
\qquad G'(x) = g(x)\\
&=c(b-2h(c)+a)-(G(a)+G(b)-2G(h(c)))\\
\end{array}
$
We want to minimize this.
$\begin{array}\\
r(c)
&=c(b-2h(c)+a)-(G(a)+G(b)-2G(h(c)))\\
&=c(b+a)-2ch(c)-(G(a)+G(b)-2G(h(c)))\\
\text{so}\\
r'(c)
&=(b+a)-2(h(c)+ch'(c))+2h'(c)G'(h(c))\\
&=(b+a)-2(h(c)+ch'(c))+2h'(c)g(h(c))\\
&=(b+a)-2(h(c)+ch'(c))+2h'(c)c\\
&=(b+a)-2h(c)\\
\end{array}
$
We want
$r'(c) = 0$,
so
$b+a
= 2h(c)
=2g^{-1}(c)
$
so
$c
=g(\frac{b+a}{2})
$.
For the problem,
$g(c) = \dfrac1{c}$
so
$c
=\dfrac{2}{b+a}
$.
Then,
since
$G(c) = \ln(c)$
and $ch(c) = 1$,
$\begin{array}\\
r(c)
&=c(b+a)-2ch(c)-(G(a)+G(b)-2G(h(c)))\\
&=\dfrac{2}{b+a}(b+a)-2-(\ln(a)+\ln(b)-2\ln(\dfrac{b+a}{2}))\\
&=-(\ln(a)+\ln(b)-2\ln(\dfrac{b+a}{2}))\\
&=\ln(\dfrac{(b+a)^2}{4ab})\\
\end{array}
$
If $a=1, b=3$
this is
$\ln(\frac{16}{12})
=\ln(\frac{4}{3})
=\ln(4)-\ln(3)
$.
