# $\lim_{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} \sin(\alpha x)\cos(\beta x) \,dx$

I have to compute this limit:

Suppose that $$|\alpha| \neq |\beta |$$, compute the following limit: \begin{align} \lim_{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} \sin(\alpha x)\cos(\beta x) \, dx \end{align}

What I've done is this: \begin{align} & \int_0^T \sin(\alpha x)\cos(\beta x) \, dx \\[8pt] = {} & \frac{1}{2(\alpha-\beta)} + \frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \end{align}

Then,

\begin{align} \lim_{T \rightarrow \infty}\frac{1}{T}\int_0^T \sin(\alpha x) \cos(\beta x) \,dx=\lim_{T \rightarrow \infty}\frac{1}{T}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ] \end{align}

But I'm not sure how can I conclude. What I think that I can do is this:

\begin{align} \underbrace{\lim_{T \rightarrow \infty}\frac{1}{T}}_{=0}\cdot \underbrace{\lim_{T \rightarrow \infty}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]}_{\text{I don't know how to compute it}} \end{align}

So intuitively I think this must be zero. How can I conclude? Is there any other way to solve it? I really appreciate your help.

• Just show that the right hand limit is bounded by a constant M, then any limit converging to 0 times a bounded sequence will also converge to 0. Commented Jan 4, 2021 at 19:10
• Thank you very much! But, can you give me please any hint about which could be that bound? @AndrewShedlock Commented Jan 4, 2021 at 22:17
• well $\alpha, \beta$ are fixed constants and cosine is bounded between $[-1,1]$. Commented Jan 5, 2021 at 20:58

## 1 Answer

Your last step doesn't work because $$\lim_{T \rightarrow \infty}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]$$ does not exist. It doesn't exist becuase the cosines oscillate rather than approaching a limit.

But you can compute $$\lim_{T \rightarrow \infty}\frac{1}{T}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]$$ by squeezing. You have $$\lim_{T\to\infty} \frac 1 T \Big[\cdots\cdots\cdots \Big]$$ where the expression in the $$\displaystyle\Big[\cdots\text{ large square brackets }\cdots\Big]$$ remains bounded, i.e. remains between two identifiable bounds, as $$T$$ changes. Thus you seek the limit of an expression that is squeezed between two things that approach $$0.$$

• Thank you very much! But, can you give me please any hint about which could be those bounds? @Michael Hardy Commented Jan 4, 2021 at 22:16
• I got it! With the fact that $-1 \leq cos(x) \leq 1$. Thank you so much!! Commented Jan 5, 2021 at 0:55
• @luisegf : Correct. (But you should write $\cos(x)$ or $\cos x$ rather than $cos(x). \qquad$ Commented Jan 5, 2021 at 5:57