# How to apply squeeze theorem to this limit.

I'm trying to solve $$\int_0^∞ e^{-x} \cos(x)\,dx$$

It is not hard to find that $$\int e^{-x} \cos(x)=\frac{1}{2}(e^{-x} \sin(x)-e^{-x} \cos(x))+C$$

From all this follows that

$$\lim_{t\to\infty}\int_0^te^{-x} \cos(x) \, dx = \frac{1}{2}\lim_{t\to\infty}(e^{-t} \sin(t) - e^{-t} \cos(t))+\frac{1}{2}$$

Notice that I have simplified already a lot the expression we are taking the limit of.

I have not been able to find this limit; a collegue student told me that I had to use the squeeze theorem, but I do not find how nor where. Any guides on how the theorem can help with this limit?

• Hint: $$\lvert\sin(t)-\cos(t)\rvert\le 2$$ so $$\lim_{t\to\infty} \exp(-t) \cdot (\sin(t)-\cos(t))=0$$ (In fact, we have the stronger bound $$\lvert\sin(t)-\cos(t)\rvert\le \sqrt 2$$ but we don't need it) Jan 11, 2020 at 0:01
• I try to make answers as simple as they can be, and I think what I've posted below is simpler than the other answers. Jan 11, 2020 at 0:22

The values of trigonometric functions oscillate between $$1$$ and $$-1$$.

$$\lim_{t \to \infty} \int_{0}^{t} e^{-x}\cos(x)\,dx = \frac{1}{2} \lim_{t \to \infty}(e^{-t}(1) - e^{-t}(-1)) + \frac{1}{2}.$$

We don’t know what value our trigonometric functions take at infinity but they can’t take any larger value clearly. So we set $$\cos$$ to $$-1$$ and $$\sin$$ to $$1$$ as it will give $$\frac{1}{2}(2e^{-t})+\frac{1}{2}$$.

And our lower bound is:

$$\lim_{t \to \infty} \int_{0}^{t} e^{-x}\cos(x)\,dx = \frac{1}{2} \lim_{t \to \infty}(e^{-t}(-1) - e^{-t}(1)) + \frac{1}{2}.$$

Which gives us $$\frac{1}{2}(-2e^{-t}) + \frac{1}{2}$$.

It’s clear that $$e^{-t}$$ goes to zero and so our limit is ‘squeezed’ to $$\frac{1}{2}$$ from both sides.

• The operations and expressions accompanied by a clear explanation of the logic and procedure. Thank you for the great response! Jan 11, 2020 at 0:23
• Glad I could help Jan 11, 2020 at 0:26

It is easiest to compute this integral using the identity $$\cos x = \frac12(e^{ix}+e^{-ix})$$. From this we have \begin{align} \int_0^\infty e^{-x}\cos x\ \mathsf dx &= \frac12\int_0^\infty e^{-x}(e^{ix}+e^{-ix})\ \mathsf dx\\ &= \frac12\int_0^\infty \left((e^{(i-1)x}+e^{-(i+1)x}\right)\ \mathsf dx\\ &=\frac12\left[\frac1{i-1}e^{(i-1)x} - \frac1{i+1}e^{-(i+1)x} \right]_0^\infty\\ &= \frac12\left(\frac1{i+1} - \frac1{i-1} \right)\\ &=\frac12\left(\frac1{1+i} + \frac1{1-i}\right)\\ &=\frac12\left(\frac{1-i + 1 + i}{(1+i)(1-i)} \right)\\ &= \frac12\left(\frac2{1-i^2} \right)\\ &= \frac12. \end{align}

• Or you could write it as$$\Re\int_0^\infty e^{-(1-i)x}dx=\Re\frac{1}{1-i}=\Re\frac{1+i}{2}=\frac12.$$
– J.G.
Jan 11, 2020 at 11:32

$$\because\sin x-\cos x=\sqrt 2(\cos\frac{\pi}{4}\sin x-\sin\frac{\pi}{4}\cos x)=\sqrt 2\sin (x-\frac{\pi}{4})$$ $$\therefore\lim_{t\rightarrow +\infty}|e^{-t}\sin t-e^{-t}\cos t|= \lim_{t\rightarrow +\infty}|\sqrt 2e^{-t}\sin (t-\frac{\pi}{4})|\leqslant \lim_{t\rightarrow +\infty}|\sqrt 2e^{-t}|=0$$

$$\therefore\lim_{t\rightarrow +\infty}(e^{-t}\sin t-e^{-t}\cos t)=0$$

$$\therefore\frac{1}{2}\lim_{t\rightarrow +\infty}(e^{-t}\sin t-e^{-t}\cos t)+\frac{1}{2}=\frac{1}{2}$$

\begin{align} & \lim_{t\to\infty} \left( e^{-t} \sin t - e^{-t} \cos t \right) \\[8pt] = {} & \left( \lim_{t\to\infty} e^{-t}\sin t \right) - \left( \lim_{t\to\infty} e^{-t} \cos t \right) \\ & \text{provided these last two limits exist} \\ & \text{(and are finite).} \end{align}

Then, since $$-1 \le \sin t \le 1,$$ we have $$-e^{-t} \le e^{-t}\sin t \le e^{-t}.$$ The first and third expressions above both approach $$0$$ as $$t\to\infty.$$

(And the one involving the cosine is done the same way.)