New Integral Corner Collection: Complex Analysis to the Rescue!

My latest addition the Integral Corner page is a collection of neat results that can be found using methods from Complex Analysis. Deriving these results by any other method isn’t particularly clear so I’ve called the collection Complex Analysis to the Rescue! Let’s visit the first few results.

The first of the collection asserts that,

1) For a,b,c \in \mathbb{R} satisfying |a|>|b|+|c|>0, and given n \in \mathbb{Z}_{\geq 0}, there hold

\displaystyle \int_0^{2\pi}\frac{\cos(nx)}{a+b\cos(x)+c\sin(x)}dx=2\pi \lambda (a,b,c,n) \cos(n\phi)

and

\displaystyle \int_0^{2\pi}\frac{\sin(nx)}{a+b\cos(x)+c\sin(x)}dx=2\pi \lambda (a,b,c,n) \sin(n\phi)

where

\displaystyle \lambda (a,b,c,n):=\frac{(\text{sgn}(a))^n\left(-|a|+\sqrt{a^2-b^2-c^2}\right)^n}{\sqrt{(a^2-b^2-c^2)(b^2+c^2)^n}}

and

\displaystyle \phi:=\text{arg}(b+ic)\in[-\pi,\pi).

To derive these results one can rewrite each integral as a complex contour integral over the unit circle |z|=1 in \mathbb{C} and employ the Residue Theorem after having checked the positions of poles of the integrand relative to the unit disc (which is where the inequality involving a,b and c comes in).

A similar approach can be used to tackle the second result in my collection which is a generalisation of the above that tells more about the kind of result to expect when integrating quotients of the form

\displaystyle \frac{a_1\cos(nx)+a_2\sin(nx)+a_3}{a_4\cos(mx)+a_5\sin(mx)+a_6}

over one period, provided |a_6|>|a_5|+|a_4|>0 and n,m are nonnegative integers. It reads

2) For any a,b,c,d,e,f\in\mathbb{R} and m,n\in\mathbb{Z}_{\geq 0} such that c>|a|+|b|>0 and m\geq 1,

\displaystyle \int_{0}^{2\pi}\frac{d\cos(nx)+e\sin(nx)+f}{a\cos(mx)+b\sin(mx)+c}\mathrm{d}x=\frac{2\pi }{m\sqrt{c^2-a^2-b^2}}\sum_{k=0}^{m-1}\left(f+\sqrt{d^2+e^2}R\cos\left(\beta-n\alpha_k\right)\right)

where

\displaystyle R:=\left(\frac{c-\sqrt{c^2-a^2-b^2}}{\sqrt{a^2+b^2}}\right)^{\frac{n}{m}},\text{ } \beta:=\text{arg}(d+ie)\in(-\pi,\pi],

and, for k\in\{0,\cdots,m-1\}, we define

\displaystyle \alpha_k:=\frac{\theta+(2k+1)\pi}{m} \text{ with } \theta=\text{arg}(a+bi) \in (-\pi,\pi].

Using the first result above in 1), one can easily tackle the following third item of the collection

3) Given |a|>|b|+|c|>0 where a,b,c\in\mathbb{R} and m\in\mathbb{Z}_{\geq 0}, for all z\in\mathbb{R} there holds

\displaystyle \int_0^{2\pi}\int_0^{2\pi}\frac{(\cos(mx)-\cos(my))(\cos(my)-\cos(mz))}{(a+b\cos(x)+c\sin(x))(a+b\cos(y)+c\sin(y))}\mathrm{d}x\mathrm{d}y=\frac{2\pi^2}{a^2-b^2-c^2}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)

Consequently, for all n\in\mathbb{N} and x_{2n+1}\in\mathbb{R} we have

\displaystyle \int_0^{2\pi}\int_0^{2\pi}\cdots\int_0^{2\pi}\prod_{j=1}^{2n}\frac{\cos(mx_{j})-\cos(mx_{j+1})}{a+b\cos(x_{j})+c\sin(x_{j})}\mathrm{d}x_1\cdots\mathrm{d}x_{2n-1}\mathrm{d}x_{2n}= \frac{2^n\pi^{2n}}{(a^2-b^2-c^2)^n}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)^n.

Notice the first result in 3) does not depend on z\in\mathbb{R}, which allows us to deduce the second result by Fubini’s Theorem and induction on n. The same approach allows one to derive the following 4th result involving an odd number of terms in the product integrand instead.

4) Given |a|>|b|+|c|>0 where a,b,c\in\mathbb{R} and m,n\in\mathbb{Z}_{\geq 0}, for all x_{2n+2}\in\mathbb{R} there holds

\displaystyle \int_0^{2\pi}\int_0^{2\pi}\cdots\int_0^{2\pi}\prod_{j=1}^{2n+1}\frac{\cos(mx_{j})-\cos(mx_{j+1})}{a+b\cos(x_{j})+c\sin(x_{j})}\mathrm{d}x_1\cdots\mathrm{d}x_{2n}\mathrm{d}x_{2n+1}= \Gamma(a,b,c,n,m)\left(\lambda(a,b,c,m)\cos(m\phi)-\lambda(a,b,c,0)\cos\left(mx_{2n+2}\right)\right)

where

\displaystyle \Gamma(a,b,c,n,m):=\frac{2^{n+1}\pi^{2n+1}}{(a^2-b^2-c^2)^n}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)^n,

\displaystyle \lambda (a,b,c,w):=\frac{(\text{sgn}(a))^w\left(-|a|+\sqrt{a^2-b^2-c^2}\right)^w}{\sqrt{(a^2-b^2-c^2)(b^2+c^2)^w}},

and \phi:=\text{arg}(b+ic)\in[-\pi,\pi).

Other problems I hope to consider involve integrands of the form

\displaystyle \frac{f(a_1\cos(nx)+a_2\sin(nx)+a_3)}{a_4\cos(mx)+a_5\sin(mx)+a_6}

where f is an analytic function locally.

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