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How can I show that

$$\int_{0}^{\frac{\pi}{4}}\ln(\sqrt{\tan x}+\sqrt{\cot x} -\sqrt{2})\ dx=0$$

I saw this integral on AoPS, and one person provides a solution:

Let $$I:=\int_0^\frac{\pi}{4} \ln(\sqrt{\tan x}+\sqrt{\cot x}-\sqrt 2 ) \ dx, \ \ \ \ \ J:=\int_0^\frac{\pi}{4} \ln(\sqrt{\tan x}+\sqrt{\cot x}+\sqrt 2 ) \ dx.$$I show that $$I=0, \ \ \ \ \ J=\frac{\pi}{2}\ln 2. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$We have $$I+J=\int_0^{\pi/4}\ln(\tan x+\cot x) \ dx=\int_0^{\pi/4}(\ln 2-\ln(\sin(2x)) \ dx=\frac{1}{2}\int_0^{\pi/2}(\ln 2-\ln(\sin x)) \ dx$$$$=\frac{\pi}{4}\ln 2-\frac{1}{2}\int_0^{\pi/2}\ln(\sin x) \ dx=\frac{\pi}{2}\ln 2. \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$Next is to compute $I-J.$ To avoid the mess, as much as we can, we put $\tan x = t^2$ to get $$I=\int_0^1\ln\left(t+\frac{1}{t}-\sqrt{2}\right)\frac{2t}{t^4+1} \ dt, \ \ \ \ \ J=\int_0^1\ln\left(t+\frac{1}{t}+\sqrt{2}\right)\frac{2t}{t^4+1} \ dt$$and changing $t$ to $1/t$ gives $$I=\int_1^{\infty}\ln\left(t+\frac{1}{t}-\sqrt{2}\right)\frac{2t}{t^4+1} \ dt, \ \ \ \ \ \ J=\int_1^{\infty}\ln\left(t+\frac{1}{t}+\sqrt{2}\right)\frac{2t}{t^4+1} \ dt.$$Thus $$I=\int_0^{\infty}\ln\left(t+\frac{1}{t}-\sqrt{2}\right)\frac{t}{t^4+1} \ dt, \ \ \ \ \ \ J=\int_0^{\infty}\ln\left(t+\frac{1}{t}+\sqrt{2}\right)\frac{t}{t^4+1} \ dt$$and hence, since $\int_0^{\infty} \frac{t\ln t}{t^4+1} \ dt=0$ (just change $t$ to $1/t$ to see that), we get $$I=\int_0^{\infty}\frac{t\ln(t^2-\sqrt{2}t+1)}{t^4+1} \ dt, \ \ \ \ \ J=\int_0^{\infty}\frac{t\ln(t^2+\sqrt{2}t+1)}{t^4+1} \ dt.$$Therefore $$I-J=\int_0^{\infty}\frac{t}{t^4+1}(\ln(t^2-\sqrt{2}t+1)-\ln(t^2+\sqrt{2}t+1)) \ dt. \ \ \ \ \ \ \ \ \ \ \ (3)$$We now use integration by parts with $\frac{t}{t^4+1} \ dt=dv, \ \ \ \ln(t^2-\sqrt{2}t+1)-\ln(t^2+\sqrt{2}t+1)=u.$ Then $$v=\frac{1}{2}\tan^{-1}(t^2), \ \ \ du=\frac{2\sqrt{2}(t^2-1)}{t^4+1} \ dt$$and so $(3)$ becomes $$I-J=-\sqrt{2}\int_0^{\infty}\frac{t^2-1}{t^4+1}\tan^{-1}(t^2) \ dt=-\sqrt{2}\int_0^{\infty}\frac{t^2-1}{t^4+1}\int_0^1\frac{t^2}{s^2t^4+1} \ ds \ dt=-\sqrt{2}\int_0^1\int_0^{\infty}\frac{t^2(t^2-1)}{(t^4+1)(s^2t^4+1)} \ dt \ ds$$$$=-\sqrt{2}\int_0^1\frac{1}{1-s^2}\left(\int_0^{\infty}\frac{s^2t^2+1}{s^2t^4+1} \ dt-\int_0^{\infty}\frac{t^2+1}{t^4+1} \ dt \right)ds.$$So changing $t$ to $t/\sqrt{s}$ in $\int_0^{\infty}\frac{s^2t^2+1}{s^2t^4+1} \ dt$ gives $$I-J=-\sqrt{2}\int_0^1\frac{1}{1-s^2}\left((\sqrt{s}-1)\int_0^{\infty}\frac{t^2}{t^4+1} \ dt+\left(\frac{1}{\sqrt{s}}-1\right)\int_0^{\infty} \frac{dt}{t^4+1}\right)ds$$$$=-\frac{\pi}{2}\int_0^1\frac{1}{1-s^2}\left(\sqrt{s}+\frac{1}{\sqrt{s}}-2\right)ds=-\frac{\pi}{2}\ln 2. \ \ \ \ \ \ \ \ \ \ \ (4)$$Now $(1)$ follows from $(2)$ and $(4).$

This is an elegant solution, but I wonder if there are other ways to solve the integral.

Larry
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  • I think sub $x=\frac{\pi}{4}-y$ would help here. – Ali Shadhar Aug 07 '19 at 23:47
  • Have you tried exploiting symmetry ? Also, $$\sqrt2 = \dfrac1{\sqrt2} + \dfrac1{\sqrt2} = \sin(\pi/4) + \sin(\pi/4) = \cos(\pi/4) + \cos(\pi/4) =$$ $$= \sin(\pi/4) + \cos(\pi/4).$$ – Lucian Aug 08 '19 at 01:23
  • I don't quite see the symmetry for this problem. – Larry Aug 08 '19 at 04:14
  • It's a nice problem but I feel that upvotes are unecessay here as the OP has pasted in a solution to a question asked and answered on AoPS. –  Aug 08 '19 at 09:22
  • AoPS provides one solution, and I am seeking other methods. – Larry Aug 08 '19 at 11:49
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    @Larry I know that, however writing out somebody else's question and answer with your actual question the last few words does not, IMO, constitute a good question. –  Aug 08 '19 at 13:07
  • Ok, I will take notice of that next time. – Larry Aug 08 '19 at 20:32
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    Note that \begin{equation} I-J =\int_{-\infty}^{\infty}\dfrac{t,\ln(t^2-\sqrt{2}t+1)}{t^4+1}, dt =f(\sqrt{2})-f(0) \end{equation} where \begin{equation} f(s)= \int_{-\infty}^{\infty}\dfrac{t,\ln(t^2-st+1)}{t^4+1}, dt. \end{equation} Furthermore \begin{equation} f'(s) = \int_{-\infty}^{\infty}\dfrac{-t^2}{(t^4+1)(t^2-st+1)}, dt=\pi\left(\dfrac{1}{\sqrt{2}+\sqrt{4-s^2}}-\dfrac{1}{\sqrt{4-s^2}}\right) \end{equation} and \begin{equation} f(\sqrt{2})-f(0)=\int_{0}^{\sqrt{2}}f'(s), ds = -\dfrac{\pi\ln 2}{2}. \end{equation} – JanG Aug 09 '19 at 16:14
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    In this answer of mine, I was able to show that $\int_0^{\frac \pi 4} \ln\left(1+\tan x + \sqrt{2\tan x} \right) dx = -\frac{G}{2} +\frac{\pi \ln 2}{2}$ by complex method. Combining with $\int_0^{\frac\pi 4} \ln (\tan x) dx = -G$, we can see that $J = \frac{\pi \ln 2}{2}$ and $I = (I+J)-J = 0$. – Myunghyun Song Aug 09 '19 at 17:14
  • $@$Song Interesting technique. – JanG Aug 09 '19 at 19:17

1 Answers1

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With $\int^{\frac{\pi}{2}}_{0}\ln({\tan x})dx=0$ \begin{align} &\int_{0}^{\frac{\pi}{4}}\ln(\sqrt{\tan x}+\sqrt{\cot x} -\sqrt{2})\overset{x\to \frac\pi2-x}{dx}\\ =& \ \frac12 \int^{\frac{\pi}{2}}_{0}\ln(\sqrt{\tan x}+\sqrt{\cot x} -\sqrt{2})\ dx\\ =&\ \frac12\int^{\frac{\pi}{2}}_{0}\ln({\tan x}-\sqrt{2\tan x} +1)\ dx \\ =& \ \frac12\cdot \frac32 \int_0^\infty \frac{\ln \frac{y^2+1}4}{y^2+1}dy=0 \end{align} where the $y$-integral in the last step results from here.

Quanto
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