A Putnam Integral $\int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}.$

Solution 1:

Let $$ \mathcal{I}=\int_{2}^{4}\dfrac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}\,\mathrm{d}x $$ Now, use that $$ \int_{a}^{b}f(x)\,\mathrm{d}x\overset{(1)}{=}\int_{a}^{b}f(a+b-x)\,\mathrm{d}x $$ Then, $$ \mathcal{I}=\int_{2}^{4}\dfrac{\sqrt{\ln(3+x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}\,\mathrm{d}x $$ Add up these two integrals to get $$ 2\mathcal{I}=\int_{2}^{4}\dfrac{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}\,\mathrm{d}x $$ Thus, $$ \mathcal{I}=1 $$

In order to prove $(1)$, write the integral using another variable, say, $t$: $$ \int_{a}^{b}f(a+b-x)\,\mathrm{d}x=\int_{a}^{b}f(a+b-t)\,\mathrm{d}t $$ In the latter one, set $x=a+b-t$ and $\mathrm{d}t=-\mathrm{d}x$ and change the limits of integration to obtain $$ \begin{aligned} \int_{a}^{b}f(a+b-t)\,\mathrm{d}t&=-\int_{b}^{a}f(x)\,\mathrm{d}x\\ &=\int_{a}^{b}f(x)\,\mathrm{d}x. \end{aligned} $$