8

Problem: Show that $$\int_{0}^{\pi / 2} \ln\left(\tan x - \sqrt{2 \tan x} + 1\right)\,\mathrm{d}x = 0 $$ If possible, I would like to use regular single-variable calculus methods, with only substitutions, IBP, partial fractions and so on, which does not involve series manipulation.

Thanks.

Quanto
  • 97,352
user149844
  • 1,245
  • suspect this is false, but might work if you had $2 \sqrt {\tan x} $ instead of $ \sqrt {2 \tan x}. $ In any case, where did you get this? – Will Jagy Sep 05 '14 at 02:01
  • 1
    Wolfram and Maple say this is true. I need to prove this to compute $$ \int_{-\infty}^{\infty} \frac{\left(x^2 - 1\right)\arctan(x^2)}{1 + x^4},\mathrm{d}x $$. – user149844 Sep 05 '14 at 02:09
  • @user149844: The value to your integral in the comment is $ \ln ( 2 ) \sqrt{2}\pi $ – Mhenni Benghorbal Sep 05 '14 at 02:45
  • 1
    But how could I deduce this value? In an equivalent manner, one could show that $$\int_{0}^{\pi /2} \ln \left(\tan x + \sqrt{2 \tan x} + 1\right),\mathrm{d}x = -2\int_{0}^{\pi /2}\ln\left( \cos x\right),\mathrm{d}x = -2\int_{0}^{\pi /2} \ln\left(\sin x\right),\mathrm{d}x = \pi \ln 2 $$. Just add and subtract $ \ln\left(\tan x +\sqrt{2 \tan x} + 1\right) $ to the integrand and use logarithms properties, but I do not know if this makes the problem someway easier. – user149844 Sep 05 '14 at 04:54
  • See https://math.stackexchange.com/questions/3305736/int-0-infty-frac-ln-left1x-sqrt2x-right1x2-dx?noredirect=1 – FDP Sep 13 '22 at 10:10

1 Answers1

6

Substitute $t^2=\tan x$, together with $t^4+1=(t^2-\sqrt2 t+1)(t^2+\sqrt2 t+1) $ \begin{align} I=&\int_{0}^{\pi / 2} \ln\left(\tan x - \sqrt{2 \tan x} + 1\right)\,dx \\ =&\int_0^\infty \frac{2t\ln(t^2-\sqrt2 t+1)}{t^4+1}dt\\ =&\int_0^\infty \frac{t\ln(t^4+1)}{t^4+1}\overset{y=t^2}{dt} + \frac12\int_{-\infty}^\infty \underset{=J}{\frac{t\ln\frac{t^2-\sqrt2 t+1}{t^2+\sqrt2 t+1}}{t^4+1}}dt\tag1\\ \end{align} Further decompose the integral $J$ as follows \begin{align} J=& \ \frac1{\sqrt2}\int_{-\infty}^\infty \overset{y=\sqrt2t-1}{\frac{\ln(t^2-\sqrt2t+1)}{t^2-\sqrt2t+1}}+ \overset{y=\sqrt2t+1}{ \frac{\ln(t^2+\sqrt2t+1)}{t^2+\sqrt2t+1}}-\overset{y=\frac{t-1/t}{\sqrt2}}{\frac{(1+t^2)\ln(t^4+1)}{t^4+1}}\ dt\\ =& \ 2\int_{-\infty}^\infty \frac{\ln \frac{y^2+1}2}{y^2+1}dy - \int_{-\infty}^\infty \frac{\ln (2(y^2+1))}{y^2+1}dy =2 \int_{0}^\infty \frac{\ln \frac{y^2+1}8}{y^2+1}dy \end{align} Plug $J$ into (1) to arrive at \begin{align} I =\frac12 \int_0^\infty \frac{\ln(y^2+1)}{y^2+1}dy + \int_{0}^\infty \frac{\ln \frac{y^2+1}8}{y^2+1}dy =\ \frac32 \int_0^\infty \frac{\ln \frac{y^2+1}4}{y^2+1}dy=0 \end{align}

Quanto
  • 97,352