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I have tried this integral in multiple ways but I can not reach a solution, it is done using integration under de integral sign.

$$\int^{\pi/2}_{0}{\ln{\left(\alpha\sin^2x+\beta\cos^2x\right)}dx}\quad \alpha,\beta\gt0$$ After doing the corresponding differentiation I ended up with this expression $$\int^{\pi/2}_{0}{-\frac{\cos^2x\sin^2x}{\left(\alpha\sin^2x+\beta\sin^2x\right)^2}dx}$$

Maybe I am doing something wrong, but I can't figure out what to do next.

Isaiah
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Samuel A. Morales
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4 Answers4

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$\int^{\pi}_{0} {\ln(1+r^2 -2r\cos t)}dt=0$

\begin{align} & \int^{\frac{\pi}{2}}_{0}{\ln{\left(\alpha\sin^2x+\beta\cos^2x\right)}dx}\\ = & \>\frac12\int^{\pi}_{0}{\ln{\left(\frac{\alpha+\beta}2-\frac{\alpha-\beta}2\cos t\right)}dt}, \>\>\>\>\>t=2x\\ = & \>\frac12\int^{\pi}_{0} {\ln\left( \frac14{(\sqrt\alpha+\sqrt\beta)^2}(1+r^2 -2r\cos t)\right)}dt ,\>\>\> r=\frac{\sqrt\alpha-\sqrt\beta}{\sqrt\alpha+\sqrt\beta}\\ = & \>\pi\ln\frac{\sqrt\alpha+\sqrt\beta}2 \end{align}

Quanto
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$$I(a,b)=\int_0^{\pi/2}\ln(a\sin^2x+b\cos^2x)dx$$ $$\partial_aI=\int_0^{\pi/2}\frac{\sin^2x}{a\sin^2x+b\cos^2x}\,dx$$ $$\partial_bI=\int_0^{\pi/2}\frac{\cos^2x}{a\sin^2x+b\cos^2x}\,dx$$ now notice that: $$\begin{align} I&=\int^\beta\int^\alpha\int^{\pi/2}\frac{\sin^2x+\cos^2x}{a\sin^2x+b\cos^2x}\,dx\,da\,db\\ &=\int^\beta\int^\alpha\int_0^{\pi/2}\frac{1}{a\sin^2x+b\cos^2x}\,dx\,da\,db\\ &=\int^\beta\int^\alpha\frac{\pi}{2\sqrt{a}\sqrt{b}}\,da\,db\\ &=2\pi\sqrt{\alpha\beta} \end{align}$$

cjferes
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Henry Lee
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Let $$I(\beta)=\int^{\pi/2}_{0}{\ln{\left(\alpha\sin^2x+\beta\cos^2x\right)}dx}$$ Then $$ I'(\beta)=\int^{\pi/2}_{0}{\frac{\cos^2x}{\alpha\sin^2x+\beta\cos^2x}dx}=\int^{\pi/2}_{0}{\frac{1}{\alpha\tan^2x+\beta}dx}.$$ Under $\tan x\to x$, \begin{eqnarray} I'(\beta)&=&\int^{\infty}_{0}{\frac{1}{(1+x^2)(\alpha x^2+\beta)}dx}\\ &=&\frac{1}{\alpha-\beta}\int^{\infty}_{0}\left(\frac{\alpha}{\alpha x^2+\beta}-\frac{1}{1+x^2}\right)dx\\ &=&\frac{\pi}{2(\sqrt{\alpha\beta}+\beta)} \end{eqnarray} and hence $$ I(\beta)=\int \frac{\pi}{\sqrt{\alpha\beta}+\beta}d\beta=\pi\ln(\sqrt{\alpha}+\sqrt{\beta})+C. $$ Using $$ I(\alpha)=\int^{\pi/2}_{0}{\ln{\left(\alpha\right)}dx}=\frac{\pi}{2}\ln\alpha $$ one has $$ \pi\ln(2\sqrt{\alpha})+C=\frac{\pi}{2}\ln\alpha $$ or $$ C=-\pi\ln2. $$ So $$ I(\beta)=\pi\ln(\frac{\sqrt\alpha+\sqrt\beta}{2}). $$

Ali Olaikhan
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xpaul
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  • You have a mistake when you calculated I'(b), the integral is $\frac{\pi}{2 (\sqrt{a b}+b)}$. Then it is cancelled when integrating I'(b) because you also forgot to multiply it by 2. Your final answer is correct but some mistakes were involved – Samuel A. Morales May 27 '20 at 01:37
  • Some typos were in my answers. Thank you for pointing. – xpaul May 27 '20 at 13:02
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OP's integral is a special case of $J(a,b)=$ $\displaystyle\int_0^\infty\frac{\log\left(x^2+a^2\right)}{x^2+b^2}\,dx$.

$$\begin{align*} I(\alpha,\beta) &= \int_0^\tfrac\pi2 \log\left(\alpha \sin^2x+\beta \cos^2x\right) \, dx \\ &= \int_0^\infty \frac{\log\left(\frac{\alpha y^2 + \beta}{y^2 + 1}\right)}{y^2+\alpha} \, dy & y=\tan x\\ &= \sqrt\alpha \int_0^\infty \frac{\log\left(y^2+\beta\right)}{y^2+\alpha} \, dy - \int_0^\infty \frac{\log\left(y^2+1\right)}{y^2+1} \, dy & y\to\frac y{\sqrt\alpha} \\ &= \sqrt\alpha J\left(\sqrt\beta,\sqrt\alpha\right) - \pi \log2 \\ &= \pi \log\left(\frac{\sqrt\alpha+\sqrt\beta}2\right) \end{align*}$$

user170231
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