Math Subject GRE 1268 Question 55

One indirect approach:

Write

$$f(a,b) = \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$$

Then changing variables by $x = ku$, for some positive $k$,

$$f(a,b) = k \int_0^\infty \frac{e^{kau}-e^{kbu}}{(1+e^{kau})(1+e^{kbu})}du = k\, f(ka,kb) $$

The only$^*$ answer which obeys the relation $$f(a,b) = k \,f(ka,kb)$$ is Option E.

---

$^*$Footnote: As pointed out in the comments, Option A does follow this relation as well but is easy to rule out on other grounds.

To unpack my original thinking:

• Option A is not possible as we can make the integrand positive: for $a > b$, $f(a,b) > 0$
• Option B is not possible as $f(a,b)$ cannot be independent of $a$ and $b$, e.g., without explicitly calculating, it looks clear that $\partial_a f(a,b) \neq 0$.
• Now we're down to Options C, D or E. Since this is a question from a timed test and mathematicians are ruthlessly efficient (aka lazy), I don't want to evaluate the integral. Instead, can I find quickly some sort of scaling argument to rule out Options C and D? My 'fear' is that $f(ka,kb) = k\,f(a,b)$ which will instead rule out Option E and then I'll be stuck having to calculate the integral in order to differentiate between C and D.
• But no! Instead $f(a,b) = k\,f(ka,kb)$. Good!

The integral being considered is, and is evaluated as, the following. \begin{align} I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\ &= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\ &= \left( \frac{1}{b} - \frac{1}{a} \right) \, \int_{1}^{\infty} \frac{dt}{t(1+t)} \mbox{ where $t = e^{bx}$ in the first and $t = e^{ax}$ in the second integral } \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{1}{t} - \frac{1}{1+t} \right) \, dt \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \, \left[ \ln(t) - \ln(1+t) \right]_{1}^{p} \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \left[ \ln\left( \frac{p}{1 + p}\right) + \ln 2 \right] \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \left[ \ln\left( \frac{1}{1 + \frac{1}{p}}\right) + \ln 2 \right] \\ &= \left( \frac{1}{b} - \frac{1}{a} \right) \, \ln 2 \end{align}

Note: Originally the statement "This is valid if $a \neq b$" was given at the end of the solution. Upon reflection it is believed that the statement should have been "This is valid for $a,b \neq 0$".

Tale $a = 1$ and let $b\to 0^+.$ In the limit you get

$$\int_0^\infty\frac{e^x-1}{(1+e^x)2}\,dx.$$

That integral equals $\infty$ because the integrand has a positive limit at $\infty.$ The only answer that fits this phenomenon is E.