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.

  • 26
    $\begingroup$ Note that $e^{ax} - e^{bx} = (1 + e^{ax}) - (1 + e^{bx})$. $\endgroup$
    – Leucippus
    Aug 3, 2015 at 19:27
  • 1
    $\begingroup$ Leucippus, you have pretty much given him the solution step. After this, just expand the fraction. $\endgroup$ Aug 3, 2015 at 19:29
  • 5
    $\begingroup$ Since interchanging $a$ and $b$ has the effect of multiplying the integral by $-1$ you can quickly rule out B. The other answers can't be ruled out for that reason. $\endgroup$ Aug 3, 2015 at 19:29
  • $\begingroup$ Side note: I'm also studying for the Math GRE subject test and have not found the specific test you are referencing. Could you post a PDF of it (or private message/email it to me privately)? Always looking for more practice problems. Thanks. $\endgroup$ Aug 3, 2015 at 22:31
  • 1
    $\begingroup$ @NoseKnowsAll I think it's new since I looked for tests a few months ago and found 4 publicly available tests (0568,8767, 9367, 9768), then I went on their website and looked at their practice test and it was a new one. ets.org/s/gre/pdf/practice_book_math.pdf. If you know of any others please let me know too. $\endgroup$ Aug 4, 2015 at 2:36

6 Answers 6


One indirect approach:


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

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!
  • 4
    $\begingroup$ THis is a very nice solution. $\endgroup$ Aug 3, 2015 at 23:21
  • 1
    $\begingroup$ Why doesn't Option A obey that relation? $\endgroup$
    – Théophile
    Aug 4, 2015 at 13:08
  • 1
    $\begingroup$ @Theophile A does but you can rule out A pretty easily by taking $a>b$, then the integrand is positive and this has nonzero integral. $\endgroup$ Aug 4, 2015 at 13:49
  • $\begingroup$ I'm making solution videos for this test and cited the clever solution that Simon S mentioned above as an alternative solutions. I gave you props in the video, but figured you'd never see that, so wanted to add a comment here. $\endgroup$
    – Brian
    Jun 28, 2016 at 4:57
  • $\begingroup$ @Brian Thanks Brian. $\endgroup$
    – Simon S
    Jun 29, 2016 at 9:35

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 $a,b \neq 0$".

  • 3
    $\begingroup$ Thanks. I can't believe I didn't see that... I have some kind of integral phobia... bad memories from calculus. $\endgroup$ Aug 3, 2015 at 19:55
  • 3
    $\begingroup$ @user3281410 Do not take that "phobia" with you when taking the GRE. $\endgroup$
    – Leucippus
    Aug 3, 2015 at 19:58
  • $\begingroup$ @kobe Thanks for the corrections. $\endgroup$
    – Leucippus
    Aug 3, 2015 at 23:34
  • $\begingroup$ Actually, to answer this question, it is not necessary to do calculation. When you take the exam, you do not have enough time to do calculation. $\endgroup$
    – xpaul
    Aug 6, 2015 at 13:14
  • $\begingroup$ @Leucippus I don't see why $a \not= b$ is necessary; the integral would be $0$ if $a=b$. We just need that $a,b > 0$. $\endgroup$
    – Cookie
    Aug 8, 2016 at 3:58

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


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

  • 1
    $\begingroup$ This was what I assumed I should do when I was trying the test, but I couldn't justify switching the limit and the integral so I moved on the next one and didn't have time to come back. $\endgroup$ Aug 3, 2015 at 19:50

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}


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

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

  • $\begingroup$ You accidentally wrote the series and integral in the same order, twice. The first step should be the other way around. $\endgroup$ Aug 8, 2016 at 2:05
  • $\begingroup$ @symplectomorphic Fixed, thank you. $\endgroup$ Aug 8, 2016 at 7:22
  • $\begingroup$ This is a great method $\endgroup$ Jul 1, 2017 at 15:59

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.