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.

  • 27
    $\begingroup$ 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
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    $\begingroup$ Why doesn't Option A obey that relation? $\endgroup$
    – Théophile
    Aug 4, 2015 at 13:08
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    $\begingroup$ @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
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    $\begingroup$ 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 $a,b \neq 0$".

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    $\begingroup$ 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).


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