Here’s an integral that could be an easy contender for the final round at the MIT Integration
Bee or a Putnam Exam. It’s so hard that I would call it the integral from hell. First
you have to use some algebraic chicanery to get this integral in a more manageable form.
After that you have two wend your way through two separate integration techniques, and evaluate
the limit at the end. As always, you’re welcome to give this problem a try – if
you dare – and when you’re ready join me for the solution. The first little trick
is to add one and then subtract one in the numerator like this. The purpose of that trick
was to split the integrand into two separate fractions. Of course now the 1 + e^ax cancels
on the left and the 1 + e^bx cancels on the right which simplifies rather nicely to this.
So that was our first major hurdle: to get this into a form that is more user friendly
for integration techniques. Since this is an improper integral we’ll need to replace
infinity with a dummy variable, say t, and rewrite as a limit. Fortunately, since both
integrands are nearly identical with the exception of b in the first exponent and a in the second,
each one can be treated as a separate integral using the same integration
techniques. Let’s begin with a substitution. We’ll let u = e^bx and v = e^ax. Then du
will equal b times e^bx dx or bu dx, and dv will be a times e^ax dx or av dx. Finally,
we need to adjust our integration limits to go from 1 to e^bt for the left integral and
1 to e^at for the right integral. When we insert all these substitutions, we’ll get
the following new integrands in terms of u and v respectively. Now we’re ready for
everyone’s favorite integration technique: partial fractions. In this case, they’re
pretty easy to set up to solve for the coefficients. I’m only going to focus on the u integrand
since we’ll get the same result with the v integrand. We’ll set it equal to A over
u plus B over 1 + u where A and B are the coefficients we need to solve for. Let’s
combine fractions on the right side. And now we can set the numerators equal to each other
and rearrange like this: Now we just equate the constants on the left with the constants
on the right and the u terms on the left with the u terms on the right. From inspection
we can easily see that A=1 and B=-1. So now our integral looks like this. Since the integral
of 1 over u is ln u, 1 over 1 + u is ln(1+u), 1 over v is ln v and 1 over 1 + v is ln (1+v),
we can finally get rid of the integrals and rewrite so that we can evaluate using the
limits of integration. Let’s use the following handy log property to simplify these expressions
even further. Plugging in the limits of integration we’ll get the following. As t tends to infinity,
the expressions in parentheses clearly go to 1, which means we’re left with just a
little bit of simplifying. Natural log 1 is zero and natural log ½ is negative natural
log 2. Now let’s just factor out ln2. And we can write this expression as a single fraction.
BAM! There’s the answer. It goes without saying that if you got this correct you know
your integration. Did you do it differently? Let me know in the comments below and as always,
thanks for watching and see you soon!