Proving that $\int_{0}^{\infty}\frac{1-\cos{y}}{y^{2}(|\log{\lambda}-\log{y}|+1)}dy<\infty$

Question

During a proof in the book I'm reading, there is an integral that is not clear to me is finite; It's the next: $$\int_{0}^{\infty}\frac{1-\cos{y}}{y^{2}(|\log{\lambda}-\log{y}|+1)}dy$$ where $\lambda>0.$

I've tried to find a function which dominates the integrand with infinite integral but I don't get it. Why this integral is finite?

Any kind of help is thanked in advanced.

Answer 1

We have \begin{align*} \left|\dfrac{1-\cos y}{y^{2}}\right|<1,~~~~y\in(0,\epsilon_{0}] \end{align*} for a small $\epsilon_{0}>0$.

For small enough $0<y<\min\{\epsilon_{0},\lambda\}$, one has $\lambda/y>1$, so $|\log\lambda-\log y|+1=\log(\lambda/y)+1>1$, then \begin{align*} \dfrac{1}{|\log\lambda-\log y|+1}<1, \end{align*} so we have controlled that \begin{align*} \int_{0}^{\min\{\epsilon_{0},\lambda\}}\dfrac{|1-\cos y|}{y^{2}(|\log\lambda-\log y|+1)}dy<\infty. \end{align*}

On the other hand, \begin{align*} \int_{1}^{\infty}\dfrac{|1-\cos y|}{y^{2}(|\log\lambda-\log y|+1)}dy\leq 2\int_{1}^{\infty}\dfrac{1}{y^{2}}dy<\infty. \end{align*}

Answer 2

$$\frac{1}{2}\int_{0}^\infty\frac{1-\cos(y)}{y^2(|\log\lambda-\log y|+1)}dy<\int_{1}^\infty\frac{1}{y^2}dy+\int_{0}^1\frac{1-\cos(y)}{y^2(|\log\lambda-\log y|+1)}dy$$

Since $1-\cos(y)$ has a max of $2$ and $|\log\lambda-\log y|$ is always positive. The second integral on the right is taken over a finite subset of the real numbers and at $y=0$ it can be shown that the limit exists using L'Hopitals Rule.