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For $a,b>0$, show that

$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\pi\ln\left(\frac{a+b}{2}\right)$$ I came across to this problem in the book Table of integrals, series and product. So here is my try to prove the closed form.

For $a>b$ then, we can write the $$ a^2\cos^2x +b^2\sin^2x =b^2\cos^2x+b^2\sin^2x + k\cos^2x = b^2 +k\cos^2x $$ where $ k= a^2-b^2$ and similarly, for $ b>a$ we can write $a^2\cos^2x +b^2\sin^2x = a^2+l\sin^2x$ where $l= b^2-a^2$. Hence we have $$ \int_0^{\frac{\pi}{2}}\ln(b^2+k\cos^2x )dx=\\ \int_0^{\frac{\pi}{2}}\left(\ln(b^2) + \ln\left(1+\frac{k}{b^2}\cos^2x\right)\right)dx\cdots(1)$$ and $$\int_0^{\frac{\pi}{2}}\ln(a^2+l\sin^2x )dx \\= \int_0^{\frac{\pi}{2}}\left(\ln(a^2) + \ln\left(1+\frac{l}{a^2}\sin^2x\right)\right)dx, \; \; b>a$$ Since $\left| \frac{k}{b^2}\cos^2 x\right|\leq 1$ so we use the Machlaurin series of $\ln(1+x)$ for $|x|<1$ giving us $$ \pi \ln(b)+\sum_{m\geq 1} \frac{(-1)^{m-1}}{m}\left(\frac{k}{4b^2}\right)^m\int_0^{\frac{\pi}{2}}\cos^{2m}xdx$$ The latter integral is well know result know as Wallis integral thus gives us $$ \frac{\pi}{2}\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\cdots(2)$$ Recalling the generating function of Central binomial coefficients for $|y| <\frac{1}{4}$, ie $$\sum_{m\geq 1} {2m\choose m} y^{m}=\frac{1}{\sqrt{1-4y}}-1.$$ Dividing by $y$ and an integrating from $0$ to $z$ we have $$ \sum_{m\geq 1}\frac{1}{m}{2m\choose m} z^{m} =-2\ln\left(\frac{\sqrt{1-4z}+1}{2}\right) $$ multiplying thoroughly by $-1$ and set $z=-\frac{k}{4b^2}=-\frac{a^2-b^2}{4b^2}$ gives us $$\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\\ = 2\ln\left(\frac{{a+b}}{2}\right)-2\ln b $$ From $(1)$ and $(2)$ we have $$\pi\ln(b)+\pi\ln\left(\frac{a+b}{2}\right)-\pi\ln b \\= \pi\ln\left(\frac{a+b}{2}\right)$$ Similarly, for the case of $b>a$ we replace $ \cos^2x$ by $\sin^2x$, $k$ by $ l$. We obtained the same desired result. Moreover, we can also have the following result.

$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\frac{1}{2}\int_0^{\pi}\ln(a^2\cos^2x+b^2\sin^2x)dx=\pi\ln\left(\frac{a+b}{2}\right)$$

Now I'm wishing to know other different approaches/proofs for the aforementioned integral.

Thank you.

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  • $\begingroup$ I've posted a new result that relies on contour integration. Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Jul 1 at 6:38
  • $\begingroup$ @Mark Viola, I could easily get your ideas till you move into complex plane portion afterward I dont get much since I'm not familiar with contour integration however, thank you for hint you left here to evaluate the integral 3. :) Indeed appreciable. $\endgroup$ – Naren Naruto Jul 1 at 7:03
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Let

$$ I(a)=\int_{0}^{\frac{\pi}{2}} \ln(a^2 \cos^2 x+b^2 \sin^2 x) dx $$

Thus

$$ I(b)=\int_{0}^{\frac{\pi}{2}} \ln(b^2 \cos^2 x+b^2 \sin^2 x) dx=\pi \ln b $$

\begin{align} I'(a)&=\int_{0}^{\frac{\pi}{2}} \frac{2a\cos^2 x}{a^2 \cos^2 x+b^2 \sin^2 x}dx \\ &=2a \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{(a^2+b^2 \tan^2 x)(1+\tan^2 x)} \\ &=\frac{\pi}{a+b} \\ \end{align}

Thus

$$ \int_b^a I'(a) da=\int_b^a \frac{\pi}{a+b} da $$

$$ I(a)-I(b)=\pi (\ln(a+b)-\ln(2b)) $$

$$ I(a)=\pi \ln \frac{a+b}{2} $$

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  • $\begingroup$ I've posted a new result that relies on contour integration. Let me know your thoughts. ;-) $\endgroup$ – Mark Viola Jul 1 at 6:39
  • $\begingroup$ @Mark Viola,Sorry,I'm not familiar with contour integration $\endgroup$ – Eeyore Ho Jul 1 at 9:35
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I thought it might be instructive to present an approach that relies on contour integration in the complex plane. To that end, we proceed.



MODIFYING THE INTEGRAL

Let $I(a,b)$ be defined by the integral

$$I(a,b)=\int_0^{\pi/2}\log(a^2\cos^2(x)+b^2\sin^2(x))\,dx\tag1$$

Using the double angle formulas, $\sin^2(x)=\frac{1-\cos(2x)}{2}$ and $\cos^2(x)=\frac{1+\cos(2x)}{2}$, in $(1)$ we obtain

$$\begin{align} I(a,b)&=\int_0^{\pi/2}\log\left(\frac{a^2+b^2}2+\frac{a^2-b^2}{2}\cos(2x)\right)\,dx\tag2\\\\ &=\frac\pi2 \log\left(\frac{a^2+b^2}{2}\right)+\frac14 \int_{-\pi}^{\pi} \log\left(1+\alpha\cos(x)\right)\,dx\tag3 \end{align}$$

In going from $(2)$ to $(3)$ we enforced the substitution $x\mapsto x/2$ and exploited the evenness of the integrand.

From here, we could use Feynman's trick, differentiate under the integral (tangent half-angle or contour integration), evaluate the integral of the derivative, and finish by integrating. Instead, we will apply contour integration to directly evaluate the integral.



MOVING TO THE COMPLEX PLANE

Next, we transform the integration on the right-hand side of $(3)$ to the complex plane through the substitution $z=e^{ix}$ to obtain
$$\begin{align} \int_{-\pi}^{\pi} \log\left(1+\alpha\cos(x)\right)\,dx&=\oint_{|z|=1}\log\left(\frac{\alpha(z^2+2z/\alpha+1)}{2z}\right)\,\frac1{iz}\,dz\tag4 \end{align}$$

where $-1<\alpha =\frac{a^2-b^2}{a^2+b^2}<1$.

We select the principal branch of the logarithm with branch cuts emanating from branch points, extending along the negative real axis, and terminating at $z=-\infty$.

Denote the roots of $z^2+2z/\alpha+1=0$ by $r_1$ and $r_2$.

We will assume that $a<b$ so that $-1<r_1=\frac{a-b}{a+b}<0$ lies inside the unit circle $|z|=1$ while $r_2=\frac{a+b}{a-b}<-1$ lies outside the unit circle.

Then, we have from $(4)$

$$\begin{align} \oint_{|z|=1}\log\left(\frac{\alpha(z^2+2z/\alpha+1)}{2z}\right)\,\frac1{iz}\,dz&=\oint_{|z|=1} \frac{\log\left(-\alpha/2\right)}{iz}\,dz-\oint_{|z|=1} \frac{\log\left(z\right)}{iz}\,dz\\\\ &+\oint_{|z|=1} \frac{\log\left(z-r_1\right)}{iz}\,dz+\oint_{|z|=1} \frac{\log\left(z-r_2\right)}{iz}\,dz\tag5 \end{align}$$

where the negative sign on $\alpha$ is required to ensure that the sum of the arguments of the logarithms is $0$ on the unit circle.



EVALUATING THE INTEGRALS IN $(5)$


From the residue theorem, we see that

$$\oint_{|z|=1} \frac{\log\left(\alpha/2\right)}{iz}\,dz=2\pi \log\left(\alpha/2\right)\tag6$$


From Cauchy's Integral Theorem, we have

$$\begin{align}\oint_{|z|=1}\frac{\log(z)}{z}\,dz&=\lim_{\varepsilon\to0}\left(\int_{-1}^{-\varepsilon}\frac{(\log(-x)-i\pi)-(\log(-x)+i\pi)}{ix}\,dx+\int_{-\pi}^\pi \frac{\log(\varepsilon e^{i\phi})}{i\varepsilon e^{i\phi}}\,d\phi\right)\\\\ &=0\tag7 \end{align}$$


From the residue theorem, we find that

$$\begin{align} \oint_{|z|=1}\frac{\log(z-r_2)}{iz}\,dz&=2\pi \log(-r_2)\tag8 \end{align}$$


Finally, from Cauchy's Integral Theorem, we have

$$\begin{align} \oint_{|z|=1}\frac{\log(z-r_1)}{iz}\,dz&=2\pi \log(-r_1)\\\\ &+\lim_{\varepsilon\to0}\left(\int_{-1}^{r_1-\varepsilon}\frac{(\log(-(x-r_1)) -i\pi)-(\log(-(x-r_1))+i\pi)}{ix}\,dx\\\\+\int_{-\pi}^{\pi} \frac{\log(\varepsilon e^{i\phi})}{i(r_1+\varepsilon e^{i\phi})}\,i\varepsilon e^{i\phi}\,d\phi\right)\\\\ &=0\tag9 \end{align}$$



PUTTING IT ALL TOGETHER

Using $(6)-(9)$, we find that

$$\begin{align} I(a,b)&=\frac\pi2 \log\left(\frac{a^2+b^2}{2}\right)+\frac\pi2 \log(-\alpha/2)+\frac\pi2\log(-r_2)\\\\ &=\pi\log\left(\frac{a+b}{2}\right) \end{align}$$

which agrees with the expected result!

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  • $\begingroup$ @narennarufo Please let me know how I can improve my answer. I really want to improve my answer. $\endgroup$ – Mark Viola 2 days ago

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