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Let $I(a,b):= \int_0^{\pi/2}\ln(a^2\cos^2x+b^2\sin^2x)\,dx$

Calculate $I(a,b)$.

My attempt:

Define function $F(a,b,x)$ as following $\frac{\partial f}{\partial a}F(a,b,x)=\ln(a^2\cos^2x+b^2\sin^2x)$

Then $I(a,b):= \int_0^{\pi/2}\ln(a^2\cos^2x+b^2\sin^2x)dx= \int_0^{\pi/2}\frac{\partial f}{\partial a}F(a,b,x)= \frac{\partial f}{\partial a}\int_0^{\pi/2}F(a,b,x)$

The last equality is done by using Leibniz integral rule, I must prove that the function is well defined in a 3-dimension cube.

Calculating F:

$F(a,b,x)=\int \ln(a^2\cos^2x+b^2\sin^2x)\,da$

Using the integral by parts, $u'=1, u=a, v=\ln(a^2\cos^2x+b^2\sin^2x), v'=\frac{2a\cos^2x}{a^2\cos^2x+b^2\sin^2x}$

$F(a,b,x)=uv-v'u=a\ln(a^2\cos^2x+b^2\sin^2x)\space-\space \int\frac{2a^2\cos^2x}{a^2\cos^2x+b^2\sin^2x}\,da= a{\ln(a^2\cos^2x+b^2\sin^2x)} \space-\space \int\frac{2a^2\cos^2x}{a^2\cos^2x+b^2\sin^2x}da= a{\ln(a^2\cos^2x+b^2\sin^2x)}\space-\space \int\frac{2a^2\cos^2x+b^2\sin^2x-b^2\sin^2x}{a^2\cos^2x+b^2\sin^2x}da= a{\ln(a^2\cos^2x+b^2\sin^2x)}-\int 1-\frac{b^2\sin^2x}{(a^2\cos^2x+b^2\sin^2x)}da$

And I am stuck.. I don't know how to handle that integral. I am not even sure I am solving it correctly..

Any tips?

Olivier Oloa
  • 120,989
Rab
  • 1,176
  • The last integral is just an arctan integral, since you are integrating with respect to $a$. – Simply Beautiful Art Mar 07 '17 at 01:31
  • since $cos^2=1-sin^2$, it's enough to calculate $\int \ln(1+C\sin (x))dx$, for $C>0$. I put this on wolfram and gave me a primitive in function of the polylogarithm function. So, i think your integral does not have elementary primitive. – Veridian Dynamics Mar 07 '17 at 01:31
  • @Basti but "we" dont knows if this answer can be written in elementary functions. The answer of Olivier Oloa seems to show that this is the case. – Masacroso Mar 07 '17 at 01:39
  • It is elementary, and even if it is not. it's a definite integral and we can calculate it using numerical ways. – Rab Mar 07 '17 at 02:03

1 Answers1

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Hint. One may set $$ f(s):=\int_0^{\pi/2}\ln(s+\sin^2 x)dx, \qquad s\geq0. $$ Then differentiating under the integral sign with respect to $s$ you get $$ \begin{align} f'(s)&=\int_0^{\pi/2}\frac1{s+\sin^2 x}dx\\\\ &=\int_0^{\infty}\frac1{s+\dfrac{t^2}{t^2+1}}\dfrac{dt}{t^2+1}\quad (t=\tan x)\\\\ &=\int_0^{\infty}\frac1{(s+1)t^2+s}dt\\\\ &=\frac{\pi}2\frac{1}{\sqrt{s(s+1)}}\\\\ &=\pi \left.\left(\ln \left(\sqrt{s}+\sqrt{s+1}\right)\right)\right|_s^{'} \end{align} $$ Thus $$ \int_0^{\pi/2}\ln(s+\sin^2 x)dx=\pi \ln \left(\sqrt{s}+\sqrt{s+1}\right)+C $$ with $C=f(0)=-\pi \ln 2$ (this one is standard) giving

$$ \int_0^{\pi/2}\ln(s+\sin^2 x)dx=\pi \ln \left(\frac{\sqrt{s}+\sqrt{s+1}}2\right), \quad s\geq0.$$

Assume without loss of generality that $b^2>a^2$, then your initial integral is obtained by writing $$ \ln\left(a^2\cos^2x+b^2\sin^2x\right)=\ln(b^2-a^2)+\ln\left(s+\sin^2x\right). $$ with $s=\dfrac{a^2}{b^2-a^2}$.

Olivier Oloa
  • 120,989
  • Can you explain how you got to the last equality? $$\dots=\pi\ln{(\sqrt(s)+\sqrt({s+1}))}$$ – Rab Mar 07 '17 at 10:19
  • @user2733996 Sure. It comes from $$\left(\ln \left(\sqrt{s}+\sqrt{s+1}\right)\right)'=\frac{\left(\sqrt{s}+\sqrt{s+1}\right)'}{\left(\sqrt{s}+\sqrt{s+1}\right)}=\frac{\frac1{2\sqrt{s}}+\frac1{2\sqrt{s+1}}}{\sqrt{s}+\sqrt{s+1}}=\frac{1}{2\sqrt{s(s+1)}}$$ Hope it is clearer now. Thanks. – Olivier Oloa Mar 07 '17 at 12:41