A friend and I have been swapping difficult integrals for the holidays to stump each other and he recently sent me this one that I haven't been able to figure out (mission accomplished, I guess :) ).

I've tried a few substitutions of the form

$$x-1 = f(t)$$

but if they cancel out one side, they won't simplify on the other because of the presence of both the exponential and the log. At best I could simplify the problem to

$$2 + \int_0^1 e^{1-\frac{1}{x^2}} + \frac{1}{\sqrt{1-\log x}}\:dx$$

by shifting the integral over to the interval $[0,1]$ to see if I could spot any patterns. The integral on the right evaluates to $1$, which is a surprisingly clean answer.

Wolfram gives a complicated looking antiderivative but one of the rules of our little game was that we would invoke no special functions beyond the standard transcendentals and hyperbolics/trig. Even if this was the intended solution, I'm not sure how to simplify the bound at $1$ with the $\operatorname{erf}$s

I suspect he had some clean trick in mind since that was the theme of the game, but I'm stumped.


Hint: Observe you have \begin{align} I=\int^2_1 e^{1-\frac{1}{(x-1)^2}}+1 +1+\frac{1}{\sqrt{1-\log(x-1)}}\ dx = \int^2_1 [f(x)+f^{-1}(x)]\ dx = 3. \end{align} Draw a picture (for any $f$)!

  • 1
    $\begingroup$ Wow that was beautiful! And I feel silly for not spotting that since my friend told me about that trick recently. Cheers! $\endgroup$ – Ninad Munshi Dec 30 '19 at 7:14
  • 1
    $\begingroup$ As an edit though, this trick only works if the functions have fixed points at the end points of the integrals. $\endgroup$ – Ninad Munshi Dec 30 '19 at 7:17

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.