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I Believe that I have the optimal solution, using the fact that: $$\int_{0}^{\pi/2}\frac{1}{1+\tan(x)^{\sqrt{2}}}=\int_{0}^{\pi/2}\frac{\cos(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}=\int_{0}^{\pi/2}\frac{\sin(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}$$

and the noticing that

$$\int_{0}^{\pi/2}\frac{\cos(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}+\int_{0}^{\pi/2}\frac{\sin(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}=\pi/2.$$

These together imply that the solution is $\pi/4$.

remark: the $\sqrt{2}$ is unimportant, for the more general $\int_{0}^{\pi/2}\frac{1}{1+\tan(x)^{\alpha}}$, my solution is invariant for any $\alpha \in {\mathbb{R}^{+}}$.

However, had I been given $$\int_{0}^{\pi/2}\frac{1}{1+\tan(x)}$$ I maybe would have actually attempted a solution by more standard methods, in order to find a closed expression for the integral. My method does not give a closed form, does anyone know a way to find one?

I was given a possible alternative route: Let $$F(\alpha)=\int_{0}^{\pi/2}\frac{1}{1+\tan(x)^{\alpha}},$$

and to differentiate with respect to $\alpha$, yet I'm not sure why this is helpful or how that kind of method "goes."

Andres Mejia
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  • This is a dumb question, but why is $\int_{0}^{\pi/2}\frac{\cos(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}+\int_{0}^{\pi/2}\frac{\sin(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}=\pi/2$ true? – Chill2Macht Jul 13 '16 at 02:50
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    @William: the two involved integrals are equal by the substitution $x\mapsto\frac{\pi}{2}-x$, and they clearly add to $\int_{0}^{\pi/2}1,dx = \frac{\pi}{2}$. For the OP: you have a wonderful solution by symmetry, why do you need a lengthier alternative? – Jack D'Aurizio Jul 13 '16 at 02:55
  • @JackD'Aurizio I think what OP is trying to ask is to find a general procedure valid for a more general class of exponents $a\in\mathbb{R}^+$ besides $\sqrt{2}$. I might be misinterpreting however. – Chill2Macht Jul 13 '16 at 03:03
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    @William: the OP is clearly well aware that the same approach works for any positive real exponent, as written in his remark. – Jack D'Aurizio Jul 13 '16 at 03:04
  • @JackD'Aurizio OK then I don't understand the question at all either – Chill2Macht Jul 13 '16 at 03:05
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    @William The OP is saying that if he saw the case where $a=1$, he wouldn't have tought of doing this trick, and might have gotten stuck when standard methods didn't work. He is asking how he could do it without resorting to such "tricks" – Ovi Jul 13 '16 at 03:16
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    @JackD'Aurizio I am proud of the solution by symmetry, and I like it a lot. I am however trying to work on different techniques of integration. It is not so much that I care about finding a better solution to this problem, but I'd like to be made aware of different ways to approach this question, particularly methods that are more extendable to other types of integrals. I hope that is reason enough to ask this question. – Andres Mejia Jul 13 '16 at 03:28
  • @labbhattacharjee 2/3 answers to that question are the same as I've posted above. The remaining answer boils down to the second part of Jack's solution. I would be glad to merge the two questions, if we can preserve Jack's answer, which I think is both different and interesting. – Andres Mejia Jul 13 '16 at 15:32

2 Answers2

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Differentiation under the integral sign is another viable alternative, but the trick is pretty the same: $$ \frac{dF}{d\alpha}=\int_{0}^{\pi/2}\left(-\log\tan x\right)\frac{\tan(x)^\alpha}{(1+\tan(x)^\alpha)^2}\,dx\\=-\frac{1}{2}\int_{0}^{\pi/2}\left(\log\tan x+\log\cot x\right)\frac{\tan(x)^\alpha}{(1+\tan(x)^\alpha)^2}\,dx =\color{red}{0}$$ gives that $F(\alpha)$ is constant over $\mathbb{R}^+$, and $$ \lim_{\alpha\to 0^+} F(\alpha) = \int_{0}^{\pi/2}\frac{dx}{2}=\color{red}{\frac{\pi}{4}}.$$

An alternative proof may go through the following lines. We have: $$ F(\alpha) = \int_{0}^{+\infty}\frac{du}{(1+u^2)(1+u^\alpha)}.$$ Through the substitution $u=v^{-1}$ we have: $$ F(\alpha)=\int_{0}^{+\infty}\frac{u^{\alpha}\,du}{(1+u^2)(1+u^{\alpha})}$$ so: $$ F(\alpha)=\frac{1}{2}\int_{0}^{+\infty}\frac{du}{1+u^2}=\frac{\pi}{4}.$$

Jack D'Aurizio
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  • Question to ensure I'm understanding, why do you examine $\lim_{\alpha \to 0_{+}}F(\alpha)$? Is it because the integral is definite, meaning that you can distribute inside the integral, and that limit evaluates to $1/2$, giving the desired answer? Can you direct me to more examples of where this method is of interest? very interesting! +1 btw – Andres Mejia Jul 13 '16 at 03:32
  • @AndresMejia: once we know that our function is constant, we just need to evaluate it at some point. $\alpha=0$ seems to grant an easy evaluation, but it is on the boundary of the domain, hence we consider the limit as $\alpha\to 0^+$ that indeed is easy to evaluate. – Jack D'Aurizio Jul 13 '16 at 03:47
  • Ah, i saw the first part of your comment, but I hadnt thought closely about $0_+$, an obvious but important point, thank you – Andres Mejia Jul 13 '16 at 03:57
  • At the risk of pestering, do you know of other instances where this method is useful? Thanks – Andres Mejia Jul 13 '16 at 03:57
  • @AndresMejia: http://math.stackexchange.com/questions/936620/how-to-find-this-double-summation is a famous exercise that involves a symmetry trick, and you may find others here on MSE by just searching for the two magical words. – Jack D'Aurizio Jul 13 '16 at 03:59
  • Ah, i was referring to the $F (\alpha) $ trick, i will look at all the symmetry arguments though, they are very cute :) much obliged – Andres Mejia Jul 13 '16 at 04:05
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    @AndresMejia: have a look at this PDF, too: https://web.williams.edu/Mathematics/lg5/Feynman.pdf – Jack D'Aurizio Jul 13 '16 at 04:07
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Let $$I= \int_{0}^{\pi/2}\frac{dx}{1+\tan x}$$ You can replace $x \rightarrow \pi/2 -x$ in the integral to get $$I=\int_{0}^{\pi/2}\frac{dx}{1+\cot x}$$ $$\implies I=\int_{0}^{\pi/2}\frac{\tan xdx}{1+\tan x}$$ Adding this to the original integral gives $$I=\frac{\pi}{4}$$

Nikunj
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  • This is not any different from my solution, $\frac{1}{1+\cot(x)}=\frac{\sin}{\cos+\sin}$... I won't downvote, but I don't think this answer adds anything to the current list of methods – Andres Mejia Jul 13 '16 at 15:30