Math Subject GRE 1268 Question 55

Question

If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$.

A: $0$

B: $1$

C: $a-b$

D: $(a-b)\log 2$

E: $\frac{a-b}{ab}\log 2$

I really don't see how to start this one, I'm not so great with integrals.

Answer 1

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.

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$^*$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!

Answer 2

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 $a,b \neq 0$".

Answer 3

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.

Answer 4

A slight variation of the accepted solution begins on line $3$:

\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}} \\ &= \int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{bx})e^{-bx}} - \int_{0}^{\infty} \frac{e^{-ax}dx}{(1 + e^{ax})e^{-ax}} \\ &=\int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{-bx})} - \int_{0}^{\infty} \frac{e^{-ax}dx}{(1 + e^{-ax})} \\ \end{align}

For the integral involving $b$, let $u=1+e^{-bx}$ so that $\frac{-1}{b}du=e^{-bx}dx$. Then $x=0 \implies u=2$, and $x \to \infty \implies u \to 1^+$. With a similar result for the integral involving $a$, and changing the order of integration, we end up with

\begin{align} I &= \left( \frac{1}{b} - \frac{1}{a} \right) \lim _{\epsilon \to 0^+}\int_{1+\epsilon}^{2}\frac{1}{u}du = \left( \frac{1}{b} - \frac{1}{a} \right) \ln 2 \ . \\ \end{align}

Answer 5

Another approach. $$\int_{0}^{\infty}\frac{e^{-bx}}{1+e^{-bx}}dx=\int_{0}^{\infty}\sum_{k\geq1}\left(-1\right)^{k+1}e^{-kbx}dx=\sum_{k\geq1}\left(-1\right)^{k+1}\int_{0}^{\infty}e^{-kbx}dx $$ $$=\frac{1}{b}\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}=\frac{1}{b}\log\left(2\right) $$ and in similar way we can calculate $\int_{0}^{\infty}\frac{e^{-ax}}{1+e^{-ax}}dx $. So

$$ \color{blue}{\int_{0}^{\infty}\frac{e^{bx}-e^{ax}}{\left(1+e^{bx}\right)\left(1+e^{ax}\right)}dx=\left(\frac{1}{b}-\frac{1}{a}\right)\log\left(2\right).}$$

Answer 6

Clearly (A) and (B) are wrong. Noting $$\lim_{a\to\infty} f(a,b) = \lim_{a\to\infty}\int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx=\int_0^\infty \frac{1}{1+e^{bx}}dx<\infty,$$ we have to choose (E).