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Let $E=(0,1]$. For every $\alpha,\beta\in\mathbb{R}$, let $f(x)=x^{\alpha}\sin{x^\beta}$. For what values of $\alpha,\beta$ is $f\in L^1(E)$?

I think I know the answer: when $\alpha>-1$ or $\alpha+\beta>-1$. I don't know how to prove that $f$ is not integrable when neither of these inequalities hold.

First suppose that $\alpha>-1$. Since $\left|\sin{x^\beta}\right|\leq 1$, we have that $\left|x^{\alpha}\sin{x^\beta}\right|\leq x^{\alpha}$. By the monotonicity of Lebesgue integration, the monotone convergence theorem, and the fact that the Lebesgue and Riemann integral of continuous functions over closed intervals are equal, we can see: $$ \int_E\left|f(x)\right|\,d\lambda\leq\int_E\left|x^\alpha\right|\,d\lambda=\int_Ex^\alpha\,d\lambda=\lim_{n\to\infty}\int_{\frac{1}{n}}^1x^\alpha\,d\lambda=\lim_{n\to\infty}\int_{\frac{1}{n}}^1x^\alpha\,dx=\lim_{n\to\infty}\frac{1^{\alpha+1}-n^{-\alpha-1}}{\alpha+1}=\frac{1}{\alpha+1}$$

Now suppose that $\alpha+\beta>-1$. So $\underset{n\to\infty}{\lim}n^{-\alpha-\beta-1}=0$. Note that $\left|\sin{u}\right|\leq u$ for all $u\geq 0$. Thus, for $x>0$ we have that $\left|\sin{x^\beta}\right|\leq x^\beta$ and so $\left| f(x)\right|\leq x^{\alpha+\beta}$ for $x\in E$. Thus, we have: $$\int_E\left|f\right|\,d\lambda\leq\int_E\left|x^{\alpha+\beta}\right|\,d\lambda=\int_Ex^{\alpha+\beta}=\lim_{n\to\infty}\int_{\frac{1}{n}}^1x^{\alpha+\beta}\,d\lambda=\lim_{n\to\infty}\int_{\frac{1}{n}}^1x^{\alpha+\beta}\,dx=\lim_{n\to\infty}\frac{1^{\alpha+\beta+1}-n^{-\alpha-\beta-1}}{\alpha+\beta+1}=\frac{1}{\alpha+\beta+1}.$$

That's all I have so far. If anyone can help me show that $f$ is not integrable when $\alpha\leq -1$ and $\alpha+\beta\leq -1$, I would be grateful. Or perhaps I am wrong and these are not the right constraints.

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1 Answer 1

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Suppose that $\alpha\leq -1$ and $\alpha+\beta\leq -1$. We will first consider the case when $\beta=0$. Then $f(x)=x^{\alpha}\sin{x^{\beta}}=x^\alpha \sin{(1)}$ for all $x\in (0,1]$. Thus, $f\in L^1(E)$ if and only if $x^{\alpha}\in L^1(E)$. Our previous argument shows us that for $\alpha\neq -1$, $\int_Ex^\alpha\,d\lambda=\lim\limits_{\epsilon\to 0}\frac{1-\epsilon^{\alpha+1}}{\alpha+1}$. This limit exists if and only if $\alpha>-1$. From the Monotone Convergence Theorem (MCT) we have $$\int_Ex^{-1}\,d\lambda=\lim\limits_{\epsilon\to 0}\int_{\epsilon}^1x^{-1}\,dx=\lim_{\epsilon\to 0}\left(\ln{(1)}-\ln{(\epsilon)}\right)=\infty.$$ Hence, if $\alpha\leq -1$, $\alpha+\beta\leq -1$, and $\beta=0$, $f\notin L^1(E)$.

Now consider the case when $\beta\neq0$. Let $u=x^{\beta}$. So $du=\beta x^{\beta-1}dx$ which means that $dx=\frac{1}{\beta}x^{1-\beta}du=\frac{1}{\beta}u^{(1/\beta)-1}du$. Further, $x^\alpha=u^{\alpha/\beta}$. So we have that $x^\alpha\sin{x^{\beta}}dx=\frac{1}{\beta}u^{((\alpha+1)/\beta)-1}\sin{u}\,du$. Let $\gamma=\frac{\alpha+1}{\beta}-1$.

If $\beta>0$, we have that:

$$ \alpha+\beta\leq -1\Rightarrow\frac{\alpha}{\beta}+1\leq-\frac{1}{\beta}\Rightarrow\frac{\alpha+1}{\beta}+1\leq 0\Rightarrow\frac{\alpha+1}{\beta}-1=\gamma\leq-2. $$

Suppose that $\beta>0$. Note that $\lim\limits_{x\to 0} x^\beta =0$ and $\lim\limits_{x\to 1}x^\beta=1$. Thus, we do not need to change the limits of integration from the improper Riemann integral $\int_0^1 x^\alpha\left|\sin{x^\beta}\right|\,dx$ to the improper Riemann integral $\int_0^1\frac{1}{\beta}u^\gamma\left|\sin{u}\right|\,du$. Since $u$ is a dummy variable, we have that $\int_0^1\frac{1}{\beta}u^\gamma\left|\sin{u}\right|\,du=\int_0^1\frac{1}{\beta}x^\gamma\left|\sin{x}\right|\,dx$. Since $x^\gamma\sin{x}=\left|x^\gamma\sin{x}\right|$ on $(0,1]$, we have from the MCT that \begin{align*} \int_Ex^\alpha\left|\sin{x^\beta}\right|\,d\lambda &=\lim_{\epsilon\to 0}\int_\epsilon^1x^\alpha\left|\sin{x^{\beta}}\right|\,d\lambda=\lim_{\epsilon\to 0}\int_\epsilon^1x^\alpha\left|\sin{x^{\beta}}\right|\,dx=\lim_{\epsilon\to 0}\int_\epsilon^1\frac{1}{\beta}x^\gamma\left|\sin{x}\right|\,dx\\ &=\lim_{\epsilon\to 0}\int_\epsilon^1\frac{1}{\beta}x^\gamma\left|\sin{x}\right|\,d\lambda=\int_E\frac{1}{\beta}x^{\gamma}\left|\sin{x}\right|\,d\lambda=\int_E\frac{1}{\beta}x^\gamma\sin{x}\,d\lambda. \end{align*} Note that $x\sin{(1)}\leq\sin{x}=\left|\sin{x}\right|$ on $(0,1]$. Thus: \begin{align*} \int_Ex^\alpha\left|\sin{x^\beta}\right|\,d\lambda &=\int_0^1\frac{1}{\beta}x^\gamma\sin{x}\,dx\geq\frac{\sin{(1)}}{\beta}\cdot\int_0^1x^{\gamma+1}\,dx\\ &=\begin{cases}\dfrac{\sin{(1)}}{\beta}\cdot\left(\lim\limits_{y\to 0}\ln{1}-\ln{y}\right)&\text{if }\gamma+1=-1\\\dfrac{\sin{(1)}}{\beta}\cdot\left(\lim\limits_{y\to 0}\dfrac{1-y^{\gamma+2}}{\gamma+2}\right)&\text{if }\gamma+1<-1\end{cases}=\infty. \end{align*} Therefore, if $\beta>0$, $\alpha\leq -1$, and $\alpha+\beta\leq -1$, $f\notin L^1(E)$.

If $\beta<0$, we have the following: $$ \alpha\leq -1\Rightarrow \alpha+1\leq 0\Rightarrow\frac{\alpha+1}{\beta}\geq 0 \Rightarrow \frac{\alpha+1}{\beta}-1=\gamma\geq -1.$$ Unlike before, when we perform the $u$-substitution, we must flip the limits of integration. So $\int_0^1x^\alpha\sin{x^\beta}\,dx=\int_{\infty}^1\frac{1}{\beta}x^\gamma\sin{x}\,dx=-\frac{1}{\beta}\int_1^\infty x^\gamma\sin{x}\,dx$. A similar argument as before gives us that $\int_0^1x^\alpha\left|\sin{x^\beta}\right|\,d\lambda=-\int_1^{\infty}\frac{1}{\beta}x^\gamma\left|\sin{x}\right|\,d\lambda$ We can see that: \begin{align*} \int_1^\infty x^\gamma \left|\sin{x}\right|\,d\lambda &\geq\int_{\pi}^\infty x^\gamma\left|\sin{x}\right|\,d\lambda=\lim_{n\to\infty}\int_\pi^{n\pi+\pi}x^{\gamma}\left|\sin{x}\right|\,d\lambda=\lim_{n\to\infty}\int_\pi^{n\pi+\pi}x^{\gamma}\left|\sin{x}\right|\,dx\\ &=\lim_{n\to\infty}\sum_{k=1}^n\int_{k\pi}^{k\pi+\pi}x^{\gamma}\left|\sin{x}\right|\,dx=\lim_{n\to\infty}\sum_{k=1}^n\int_{0}^{\pi}(x+k\pi)^{\gamma}\left|\sin{x+k\pi}\right|\,dx\\ &=\lim_{n\to\infty}\sum_{k=1}^n\int_{0}^{\pi}(x+k\pi)^{\gamma}\left|\sin{x}\right|\,dx=\lim_{n\to\infty}\sum_{k=1}^n\int_{0}^{\pi}(x+k\pi)^{\gamma}\sin{x}\,dx\\ &\geq\begin{cases}\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\int_{0}^{\pi}(\pi+k\pi)^{\gamma}\sin{x}\,dx&\text{if }\gamma\geq 0\\\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\int_{0}^{\pi}(k\pi)^{\gamma}\sin{x}\,dx&\text{if }\gamma< 0\end{cases}\\ &=\begin{cases}\pi^\gamma\lim\limits_{n\to\infty}\sum\limits_{k=1}^n(k+1)^{\gamma}\int_{0}^{\pi}\sin{x}\,dx&\text{if }\gamma\geq 0\\\pi^\gamma\lim\limits_{n\to\infty}\sum\limits_{k=1}^nk^{\gamma}\int_{0}^{\pi}\sin{x}\,dx&\text{if }\gamma< 0\end{cases}\\ &=\begin{cases}2\pi^\gamma\lim\limits_{n\to\infty}\sum\limits_{k=1}^n(k+1)^{\gamma}&\text{if }\gamma\geq 0\\2\pi^\gamma\lim\limits_{n\to\infty}\sum\limits_{k=1}^nk^{\gamma}&\text{if }\gamma< 0\end{cases}=\infty. \end{align*} Thus, $-\frac{1}{\beta}x^\gamma\sin{x}\notin L^1([1,\infty))$. By our previous argument, we have that $x^\alpha\sin{x^\beta}\notin L^1(E)$.

Thus, $f\in L^1(E)$ if and only if either $\alpha>-1$ or $\alpha+\beta>-1$.

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