Convergence of integral with trigonometry

Question

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

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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?

Answer 1

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

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

Answer 2

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.

Answer 3

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

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

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

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?

Answer 4

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.