1
$\begingroup$

Putnam 1985 B5: Evaluate $\int_{0}^\infty t^{-1/2}e^{-1985(t+t^{-1})}dt$. You may assume that $\int_{-\infty}^\infty e^{-x^2} dx=\sqrt{\pi}$.

Following the solution on https://artofproblemsolving.com/community/c1860h1093626s3_1985_putnam_b5, I arrive at the conclusion that $I=2e^{-3970}\int_{0}^\infty e^{-1985{(u-u^{-1})}^2}du$, where $u=\sqrt{t}$. Another expression for $I$ is that, $I=2e^{-3970}\int_{0}^\infty (\frac{1}{v^2})e^{-1985{(v-v^{-1})}^2}dv$, where $u=v^{-1}$. The writer then goes on to claim that, by taking the average of these two equations for $I$, and by reverting the variables back to $x$, one gets $I=e^{-3970}\int_{0}^\infty (1+\frac{1}{x^2})e^{-1985{(x-x^{-1})}^2}$. This is precisely where I lose track of the proof: I understand the idea of taking the average to get $e^{-3970}$ but I cannot understand his idea of reverting variables back to $x$ (why not $t$?) and how this process is done. Any help would be much appreciated.

Thank you.

EDIT: I have managed to now solve this problem myself case closed :)

$\endgroup$
  • $\begingroup$ $I =( I + I)/2$. And the value of definite integrals have nothing to do with the choice of alphabets. $\endgroup$ – xbh Aug 5 '18 at 3:34
2
$\begingroup$

This is a classic technique in the evaluation of definite integrals, which takes advantage of the fact that the value of a definite integral does not depend on the variable used in it, since it has a constant value. Hence, $$\int_a^b f(x)dx =\int_a^b f(u)du=\int_a^b f(ñ)dñ=...$$ And the variable (often referred to as a "dummy variable") does not matter, just as the index of a sum does not affect its value: $$\sum_{i=a}^b f(i)=\sum_{j=a}^b f(j)$$ Furthermore, if a substitution $x\to g(u)$ does not affect the bounds of the integral so that $$\int_a^b f(x)dx=\int_a^b f(g(u))g'(u)du$$ We may write tha second integral also in terms of $x$, since the $u$ is just a "dummy variable", and average the two integrals to get $$\int_a^b \frac{f(x)+f(g(x))g'(x)}{2}dx $$ Then we may (in problems like the one you proposed) make use of some clever cancellation or functional equation that causes the integrando to turn into something manageable.

$\endgroup$
  • $\begingroup$ I was aware of these properties - my question is precisely about this particular example! I can't see how he manages to reach the form with the $x$ variables following taking the average of the other two integrals... $\endgroup$ – Daniele1234 Aug 5 '18 at 2:46
  • $\begingroup$ The substitution $x \to g(u)$ does change the bounds if they are expressed in terms of $u$. Usually when we integrate with respect to a variable the bounds are on that variable. I have seen it writing $\int_a^b (stuff)dx=\int_{x=a}{x=b}(stuff')du$ to show the bounds are still on $x$. $\endgroup$ – Ross Millikan Aug 5 '18 at 3:36

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.