# Putnam 1985 B5 Integration by Substitution

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 :)

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

## 1 Answer

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.

• 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... – Daniele1234 Aug 5 '18 at 2:46
• 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 $x$. – Ross Millikan Aug 5 '18 at 3:36