Symmetrizing an integral representation of a symmetric function Here is a function of two complex numbers $a$, $b$ that I believe is symmetric under the exchange $a \leftrightarrow b$:
$$
I(a,b) = (-a+1/b)\int_0^\infty dx \frac{\ln(1+bx)}{(x+a)(x+1/b)}
$$
This integral representation does not exhibit this symmetry explicitly.  Is there a way to find a more aesthetically pleasing representation of $I(a,b)$ that is manifestly symmetric?
n.b. I don't think $\frac{1}{2}[I(a,b)+I(b,a)]$ to be aesthetically pleasing.  But if they can be brought together nicely, it's ok.
 A: Good question! The expression is clearly symmetric in $a,b$ and we can prove this from your original representation:
Multiplying on top and bottom by $b$ gives $$I(a,b) = (1-ab)\int^\infty_0 \frac{\log(1+bx)}{(x+a)(1+bx)} dx.$$ Making the substitution $x = \frac{ay}{b}$ (here I am assuming that $a,b > 0$; indeed, if $a$ and $b$ are not positive, we have singularities in the range of integration so there will be some trouble) yields \begin{align*}I(a,b) &= (1-ab) \int^\infty_0 \frac{\log(1+ay)}{\left(\tfrac{ay}{b} + a\right) (1+ay)} \frac a b dy \\ &= (1-ab) \int^\infty_0 \frac{\log(1+ay)}{(y+b)(1+ay)}dy = I(b,a). \end{align*}
To find a a representation where it is manifestly symmetric, simply make the substitution $x\mapsto ax$ (again, assuming that $a>0$). Then we see \begin{align*} I(a,b) &= (1-ab) \int^\infty_0 \frac{\log(1+abx)}{(ax+a)(1+abx)} a \,dx \\ &= (1-ab) \int^\infty_0 \frac{\log(1+abx)}{(x+1)(1+abx)} \,dx. \end{align*} Now $a$ and $b$ only appear in the product $ab$ so clearly the expression is symmetric. This also hints that $I$ should actually only be a function of one parameter.
