Why is $\int_{[0,\infty[}e^{-xy}|\sin{x}|d\lambda(x) \leq 2\int_{A}e^{-xy}\sin{x}d\lambda(x)$

Question

Let $A:= \bigcup_{k=0}^{\infty}[2k\pi,(2k+1)\pi]$.

I do not understand why the following holds, it was written in our textbook

$\int_{[0,\infty[}e^{-xy}|\sin{x}|d\lambda(x) \leq 2\int_{A}e^{-xy}\sin{x}d\lambda(x)$

Even reducing it to $\int_{[0,2\pi[}e^{-xy}|\sin{x}|d\lambda(x) \leq 2\int_{[0,2\pi[}e^{-xy}\sin{x}d\lambda(x)$

All I can note is that $e^{-xy}$ is decreasing for a fixed $y$ as $x \to \infty$

and on $[\pi,2\pi]$where $\sin{x} \leq 0$ it is hard to see why $e^{-xy}|\sin{x}|\leq2e^{-xy}\sin{x}$

Surely this cannot be true?

Answer 1

Let $I_k =[(2k+1)\pi,(2k+2)\pi]$ and $J_k =[2k\pi,(2k+1)\pi]$. Then we have $$\begin{eqnarray} \int_{I_k} e^{-xy}|\sin{x}|d\lambda(x) &=&\int_{J_k} e^{-(x+\pi)y}|\sin(x+\pi)|d\lambda(x)\\ &=&\int_{J_k} e^{-(x+\pi)y}\sin(x)d\lambda(x)\\ &\le&\int_{J_k} e^{-xy}\sin(x)d\lambda(x). \end{eqnarray}$$ Thus we have $$\begin{eqnarray} \int_{[0,\infty[}e^{-xy}|\sin{x}|d\lambda(x)&=&\sum_{k\ge 0}\left[\int_{I_k} e^{-xy}|\sin{x}|d\lambda(x)+\int_{J_k} e^{-xy}|\sin{x}|d\lambda(x)\right]\\ &\le&2\sum_{k\ge 0}\int_{J_k} e^{-xy}\sin{x}d\lambda(x)\\ &=&2\int_{\bigcup_{k\ge 0} J_k} e^{-xy}\sin{x}d\lambda(x)=2\int_{A} e^{-xy}\sin{x}d\lambda(x). \end{eqnarray}$$

Answer 2

You need to compare $I=\int_0^{\pi}e^{-xy}|sinx|d\lambda(x)$ with $J=\int_{\pi}^{2\pi}e^{-xy}|sinx|d\lambda(x)$ Because of the exponential term, $J\lt I$. (Note the same comparison holds for each interval of length $2\pi$). The right side of your inequality is made up of terms like $2I$, while the left side is made up of terms like $I+J$, so the left side is smaller for all $y\gt 0$.

I suppose if $\lambda(x)$ was weird enough, you could get equality.