# Complex Analysis approach

Hello everyone!

I encountered the following integral a while ago and solved it using Feynman's trick as follows. However, it looks like the residue theorem would be elegant for this problem. Does anyone know how that might work out?

P.S: It would be helpful if someone would check my solution as well.

$I(a,b) = \int_0^{\pi/2} \ln( a \sin^2\theta + b\cos^2\theta)\: d\theta$ Upon taking the partial derivatives wrt $a$ and $b$, we have the following: $\frac{\partial I}{\partial a} = \int_0^{\pi/2} \frac{\sin^2\theta}{a\sin^2\theta + b \cos^2\theta}d \theta, \:\:\:\: \frac{\partial I}{\partial b} = \int_0^{\pi/2} \frac{\cos^2\theta}{a\sin^2\theta + b \cos^2\theta}d \theta$ Next, observe that linear combinations of these two integrals results in integrals that are relatively simple to evaluate. In particular, $a\frac{\partial I}{\partial a} + b\frac{\partial I}{\partial b} = \int_0^{\pi/2} \frac{a\sin^2\theta + b \cos^2\theta}{a\sin^2\theta + b\cos^2\theta}d\theta = \int_0^{\pi/2} d\theta = \frac{\pi}{2}$ and upon summing the two partial derivatives normally, we have $\frac{\partial I}{\partial a} + \frac{\partial I}{\partial b} = \int_0^{\pi/2} \frac{\sin^2\theta + \cos^2\theta}{a\sin^2\theta + b\cos^2\theta}d\theta = \int_0^{\pi/2} \frac{d\theta }{a\sin^2\theta + b\cos^2\theta} = \int_0^{\pi/2} \frac{\sec^2\theta \: d\theta}{b + a\tan^2\theta}$ In order to convert this integral into the more familiar inverse tangent form, we set $\tan\theta = \sqrt{\frac{b}{a}} \tan\varphi$ so that $\sec^2\theta \: d\theta = \sqrt{\frac{b}{a}} \sec^2\varphi \: d\varphi$. For finite $a,b$ and $a\neq 0$, the limits of integration do not change. Then $\frac{\partial I}{\partial a} + \frac{\partial I}{\partial b} = \int_0^{\pi/2} \sqrt{\frac{b}{a}} \cdot \frac{\sec^2\varphi \: d\varphi}{b + a \cdot \frac{b}{a}\tan^2\varphi} = \int_0^{\pi/2} \sqrt{\frac{b}{a}} \cdot \frac{\sec^2\varphi \: d\varphi}{b + a \cdot \frac{b}{a}\tan^2\varphi} = \int_0^{\pi/2} \frac{d\varphi}{\sqrt{ab}}$ $\frac{\partial I}{\partial a} + \frac{\partial I}{\partial b} = \frac{\pi}{2\sqrt{ab}}$ Thus from (1) and (2), we have a system of partial differential equations that can easily be solved. We take $(1) - b \cdot (2)$ to get $(a-b) \frac{\partial I}{\partial a} = \frac{\pi}{2}\bigg( 1- \sqrt{\frac{b}{a}}\bigg), \text{ or equivalently }\:\:\: \frac{\partial I }{\partial a } = \frac{\pi}{2} \cdot \frac{\sqrt{a} - \sqrt{b}}{(a-b)\sqrt{a}} = \frac{\pi}{2\sqrt{a} (\sqrt{a} + \sqrt{b})}$ Upon integrating wrt $a$, we get an analytic solution for $I$, upto a constant $I(a,b) = \int \frac{\pi \: da}{2\sqrt{a}(\sqrt{a} + \sqrt{b})} = \pi \ln(\sqrt{a} + \sqrt{b}) + C$ Notice that this solution is symmetric in $a$ and $b$, just like our system of partial differential equations, so there is no need to integrate wrt $b$. We can differentiate and check that this is indeed a solution to the system of differential equations. To evaluate the constant $C$, notice that for $a=b=1$, we have $I(1,1) = \int_0^{\pi/2} \ln(\sin^2\theta + \cos^2\theta) \: d\theta = 0$ so then for our closed form, we obtain $0 = \pi\ln(2) + C, \text{ or } C = -\pi\ln(2)$ With this, we now have a general solution to this particular family of integrals, $I(a,b) = \pi\ln(\sqrt{a} + \sqrt{b}) - \pi\ln 2 = \pi \ln \frac{\sqrt{a} + \sqrt{b}}{2}$

Note by Kishan Jani
3 months, 3 weeks ago

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Here is a video by a really good math Youtuber that uses a nice approach also : https://www.youtube.com/watch?v=NAWZx77ZOpw&t=584s

Also, your solution looks good to me :)

- 2 months, 2 weeks ago

Thank you so much for the reference! :)

- 2 months, 2 weeks ago

I have derived using series manipulation, you may wish to check out here. Also you can find that second solution follows what you have requested.

- 3 weeks, 2 days ago

How did you learn all this at 17? I only recently started learning the basics on real analysis.

- 3 months, 3 weeks ago