# Convergence of integral with trigonometry

• $\displaystyle \int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}} dx$

Obviously, problem in $-\frac{\pi}{4}$ as $\lim_{x\to -\frac{\pi}{4}^+} = \infty$, integral is positive on whole segment, but I can't use any usual rules to prove it's convergence or simly divide it inside the root. I guess, i have to bound it with converging function, but can't find a good one aswell. Any thoughts?

## 4 Answers

Recall that

• $\cos x - \sin x=\sqrt 2 \sin \left(\frac{\pi}4-x\right)$

• $\cos x + \sin x=\sqrt 2 \sin \left(\frac{\pi}4+x\right)$

then

$$=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)^{\frac{1}{3}} dx =\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\sin \left(\frac{\pi}4-x\right)}{\sin \left(\frac{\pi}4+x\right)}\right)^{\frac{1}{3}} dx$$

Assume $y=x+\frac{\pi}4$ then we have

$$\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\sin \left(\frac{\pi}4-x\right)}{\sin \left(\frac{\pi}4+x\right)}\right)^{\frac{1}{3}} dx =\int_{0}^\frac{\pi}{2} \left(\frac{\sin \left(\frac{\pi}2-y\right)}{\sin y}\right)^{\frac{1}{3}} dy =\int_{0}^\frac{\pi}{2} \left(\frac{1}{\tan y}\right)^{\frac{1}{3}} dy$$

and

$$\left(\frac{1}{\tan y}\right)^{\frac{1}{3}}\sim\frac1{\sqrt[3] y}$$

The problem is indeed in $-\pi/4$.

The Taylor expansion around $-\pi/4$ for the sine: \begin{align} \sin(x) = -\frac{1}{2}\sqrt[]{2}+\frac{1}{2}\sqrt[]{2} (x+\pi/4) + O((x+\pi/4)^2) \end{align} and for the cosine: \begin{align} \cos(x) = \frac{1}{2}\sqrt[]{2}+\frac{1}{2}\sqrt[]{2} (x+\pi/4)+O((x+\pi/4)^2) \end{align} Therefore: \begin{align} \cos(x)+\sin(x) = \sqrt[]{2}(x+\pi/4) + O((x+\pi/4)^2) \end{align} So, near $-\pi/4$, the integrand behaves as: $$\frac{1}{(\sqrt[]{2}(x+\pi/4))^{1/3}}$$ That means that the original integral converges.

There are some details left for you, for example finding the convergent integral that dominates your integral, to make this argument fully rigorous.

$$I=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}}, dx$$

$1)sin(x)+cos(x)=Rcos(x-a)\therefore R=\sqrt{2}cos(x-\frac{\pi}{4})$

$$sin(x)+cos(x)=\sqrt{2}cos(x-\frac{\pi}{4})$$

do the same for $cos(x)-sin(x)$, so

$$cos(x)-sin(x)=\sqrt{2}cos(x+\frac{\pi}{4})$$

$$\therefore\frac{\cos x - \sin x}{\cos x + \sin x}=\frac{\sqrt{2}cos(x+\frac{\pi}{4})}{\sqrt{2}cos(x-\frac{\pi}{4})}=\frac{cos(x+\frac{\pi}{4})}{cos(x-\frac{\pi}{4})}$$ Can you do anything with this?

Starting from gimusi's answer, consider $$I=\int \frac{dy}{\tan^{1/3}(y)}$$ Let $$y=\tan ^{-1}\left(z^3\right)\implies dy=\frac{3 z^2}{z^6+1}\,dz$$ which make $$I=\int \frac{3 z}{z^6+1}\,dz=\int \frac{3 z}{(z^2+1)(z^4-z^2+1)}\,dz$$ Contiue the factorization of $z^4-z^2+1$.

I am sure that you can take it from here.

• Thanks for the quote! My idea was to conclude by limit comparison test but it is a nice way to explore further without asymptotics big cannon. – gimusi May 24 '18 at 13:28