exercise: prove that $$\lim\limits_{\lambda\rightarrow\pi}\int_0^\pi \sin\left(\left(\lambda+\frac{1}{2}\right)x\right)\left[\frac{2}{x}-\frac{1}{\sin\left(\frac{x}{2}\right)}\right]\, dx=0$$butI think there is a misstype in the text, it should be $$\lim\limits_{\lambda\rightarrow\infty}\int_0^\pi \sin\left(\left(\lambda+\frac{1}{2}\right)x\right)\left[\frac{2}{x}-\frac{1}{\sin\left(\frac{x}{2}\right)}\right]\, dx=0$$ because of Riemann-Lebesgue Lemma which states that if $f$ is integrable on $[a,b]$ then $$\lim\limits_{\lambda\rightarrow\infty}\int_a^b f(x)\,\sin(\lambda x)\,dx=0$$ In our case we can prove that $f(x)=\frac{2}{x}-\frac{1}{\sin\left(\frac{x}{2}\right)}$ is integrable on $[0,\pi]$ then we can apply the Riemann-Lebesgue Lemma.
If my suggestion is right????
Best Answer
Your suggestion is right and we can show that $f(x)=\frac{2}{x}-\frac{1}{\sin(\frac{x}{2})}$ is integrable in $[0,\pi]$. The singularity appears at $x=0$, so we only need to focus on the behavior of $f$ near $x=0$. We have $$f(x)=\frac{2\sin\left(\frac x2\right)-x}{x\sin\left(\frac x2\right)}=\frac{2\sin\left(\frac x2\right)-x}{x^2}\cdot\frac{x}{\sin\left(\frac x2\right)}.$$ As $x\to0$, we have $\sin\left(\frac x2\right)=\frac x2-\frac16\left(\frac x2\right)^3+O(x^4)$, hence $$\lim_{x\to0}\frac{2\sin\left(\frac x2\right)-x}{x^2}=\lim_{x\to0}\frac{2\left(\frac x2-\frac{x^3}{48}\right)-x}{x^2}=0.$$ On the other hand, we have $$\lim_{x\to0}\frac{x}{\sin\left(\frac x2\right)}=2.$$ Hence $\lim_{x\to0}f(x)=0$. In particular, if we define $f(0)=0$, then $f$ is bounded on $[0,\delta]$ for some $\delta\in(0,1)$. For $x\in [\delta, \pi]$, by the monotonicity of $x$ and $\sin(x/2)$, we have $$|f(x)|\leq\frac2x+\frac1{\sin\left(\frac x2\right)}\leq \frac2\delta+\frac1{\sin\left(\frac \delta2\right)}.$$
As a result, if we define $f(0)=0$, then $f$ is bounded on $[0,\pi]$, and hence $f$ is integrable in $[0,\pi]$. So, Riemann-Lebesgue lemma is applicable.