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)=0π/2ln(asin2θ+bcos2θ)dθI(a,b) = \int_0^{\pi/2} \ln( a \sin^2\theta + b\cos^2\theta)\: d\theta Upon taking the partial derivatives wrt aa and bb, we have the following: Ia=0π/2sin2θasin2θ+bcos2θdθ,Ib=0π/2cos2θasin2θ+bcos2θdθ\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, aIa+bIb=0π/2asin2θ+bcos2θasin2θ+bcos2θdθ=0π/2dθ=π2a\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 Ia+Ib=0π/2sin2θ+cos2θasin2θ+bcos2θdθ=0π/2dθasin2θ+bcos2θ=0π/2sec2θdθb+atan2θ \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 tanθ=batanφ\tan\theta = \sqrt{\frac{b}{a}} \tan\varphi so that sec2θdθ=basec2φdφ\sec^2\theta \: d\theta = \sqrt{\frac{b}{a}} \sec^2\varphi \: d\varphi. For finite a,ba,b and a0a\neq 0, the limits of integration do not change. Then Ia+Ib=0π/2basec2φdφb+abatan2φ=0π/2basec2φdφb+abatan2φ=0π/2dφab\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}} Ia+Ib=π2ab\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(2)(1) - b \cdot (2) to get (ab)Ia=π2(1ba), or equivalently Ia=π2ab(ab)a=π2a(a+b)(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 aa, we get an analytic solution for II, upto a constant I(a,b)=πda2a(a+b)=πln(a+b)+CI(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 aa and bb, just like our system of partial differential equations, so there is no need to integrate wrt bb. We can differentiate and check that this is indeed a solution to the system of differential equations. To evaluate the constant CC, notice that for a=b=1a=b=1, we have I(1,1)=0π/2ln(sin2θ+cos2θ)dθ=0I(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=πln(2)+C, or C=πln(2)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)=πln(a+b)πln2=πlna+b2I(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 :)

N. Aadhaar Murty - 2 months, 2 weeks ago

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Thank you so much for the reference! :)

Kishan Jani - 2 months, 2 weeks ago

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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.

Naren Bhandari - 3 weeks, 2 days ago

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

Barry Leung - 3 months, 3 weeks ago

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