Estimate from below $\int_0^\pi e^{-t}\cos nt dt$ without calculate it.

Question

Estimate from below the following integral $$\int_0^\pi e^{-t}\cos nt dt$$ without calculate it. Here $n\in\mathbb N$. Any suggestions please?

Answer 1

Here's one idea: $\cos(nt) \geq 1/2$ on $[0,\pi/3n]$, and is always $\geq -1$, so

$$\int_0^\pi e^{-t} \cos(nt) \geq \int_0^{\pi/3n} e^{-t}/2 dt - \int_{\pi/3n}^\pi e^{-t} dt.$$

I don't know if this is tight enough for your purposes, however. If I needed this a little tighter I would repeat the same idea: cut off the region where $\cos(nt) \geq 1/2$ and then bound the rest of it below (perhaps by $-1$ again). You can get tighter still by breaking up the circle into twelfths, thereby getting an estimate on $[0,\pi/6n],[\pi/6n,2 \pi/6n]$, etc. up to $[(6n-1)\pi/6n,\pi]$. Probably the most important thing for practical purposes is to capture the intuitive fact that the integral is positive.

Answer 2

$$\begin{eqnarray*} I &=& \frac{1}{n}\int_{0}^{n\pi}e^{-t/n}\cos t\,dt =\frac{1}{n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}e^{-t/n}\cos t\,dt\\&=&\frac{1}{n}\int_{0}^{\pi}\cos t\,e^{-t/n}\sum_{j=0}^{n-1}(-1)^j e^{-j\pi/n}\,dt\tag{1}\end{eqnarray*}$$ Now: $$\sum_{j=0}^{n-1}(-1)^j e^{-j\pi/n}=e^{-\pi}\frac{e^{\frac{\pi}{n}}}{1+e^{\frac{\pi}{n}}}(e^{\pi}-(-1)^n)\geq\color{red}{ \frac{1-e^{-\pi}}{2}}\tag{2}$$ so, in order to provide a lower bound, we just need to prove that: $$ J_n = \int_{0}^{\pi}e^{-t/n}\cos t \,dt \tag{3}$$ is positive for any $n\geq 1$. Easy task: $$ J_n = \int_{0}^{\pi/2}\cos t\left(e^{-t/n}-e^{(t-\pi)/n}\right)\,dt =2\,e^{-\frac{\pi}{2n}}\int_{0}^{\pi/2}\cos t \sinh\left(\frac{\frac{\pi}{2}-t}{n}\right)\,dt\tag{4}$$ hence: $$ J_n \geq \frac{2}{n}e^{-\frac{\pi}{2n}}\int_{0}^{\pi/2}\cos t\left(\frac{\pi}{2}-t\right)\,dt = \color{red}{\frac{2}{n}e^{-\frac{\pi}{2n}}}.\tag{5}$$ By putting together $(1),(2)$ and $(5)$ we have that $I$ is $\Omega\left(\frac{1}{n^2}\right)$.

Answer 3

I get that the integral goes like $\frac{\pi(1-e^{-\pi})}{n^2}$ for even $n$. Here is the annoying long derivation.

I'll assume that $n$ is even for now and write $n = 2m$.

First, split the integral into $m$ parts.

$\begin{array}\\ I(n) &=I(2m)\\ &=\int_0^\pi e^{-t}\cos n\, t\ dt\\ &=\int_0^\pi e^{-t}\cos (2mt) dt\\ &=\frac1{2m}\int_0^{2m\pi} e^{-t/(2m)}\cos t\ dt\\ &=\frac1{2m}\sum_{k=0}^{m-1}\int_{2k\pi}^{2(k+1)\pi} e^{-t/(2m)}\cos t\ dt\\ &=\frac1{2m}\sum_{k=0}^{m-1}I(m, k)\\ \end{array} $

Then, split each sub-integral into 4 parts and see how much cancellation we can get.

$\begin{array}\\ I(m, k) &=\int_{2k\pi}^{2(k+1)\pi} e^{-t/(2m)}\cos t\ dt\\ &=\int_{0}^{2\pi} e^{-(t+2k\pi)/(2m)}\cos t\ dt\\ &=e^{-(k\pi)/m}\int_{0}^{2\pi} e^{-t/(2m)}\cos t\ dt\\ &=e^{-(k\pi)/m} \left(\int_{0}^{\pi/2}+\int_{\pi/2}^{\pi}+\int_{\pi}^{3\pi/2}+\int_{3\pi/2}^{2\pi}\right) e^{-t/(2m)}\cos t\ dt\\ &=e^{-(k\pi)/m} \left(\int_{0}^{\pi/2}+\int_{\pi}^{3\pi/2}+\int_{\pi/2}^{\pi}+\int_{3\pi/2}^{2\pi}\right) e^{-t/(2m)}\cos t\ dt\\ &=e^{-(k\pi)/m} \left( \int_{0}^{\pi/2}(e^{-t/(2m)}\cos t+e^{-(t+\pi)/(2m)}\cos (t+\pi))dt +\int_{\pi/2}^{\pi}(e^{-t/(2m)}\cos t+e^{-(t+\pi)/(2m)}\cos (t+\pi))dt \right)\\ &=e^{-(k\pi)/m} \left( \int_{0}^{\pi/2}(e^{-t/(2m)}\cos t-e^{-(t+\pi)/(2m)}\cos (t))dt +\int_{\pi/2}^{\pi}(e^{-t/(2m)}\cos t-e^{-(t+\pi)/(2m)}\cos (t))dt \right)\\ &=e^{-(k\pi)/m} \left( \int_{0}^{\pi/2}((e^{-t/(2m)}-e^{-(t+\pi)/(2m)})\cos (t))dt +\int_{\pi/2}^{\pi}(e^{-t/(2m)}-e^{-(t+\pi)/(2m)})\cos (t))dt \right)\\ &=e^{-(k\pi)/m} \left( \int_{0}^{\pi/2}(e^{-t/(2m)}(1-e^{-\pi/(2m)})\cos (t))dt +\int_{\pi/2}^{\pi}(e^{-t/(2m)}(1-e^{-\pi/(2m)})\cos (t))dt \right)\\ &=e^{-(k\pi)/m} (1-e^{-\pi/(2m)})\left( \int_{0}^{\pi/2}(e^{-t/(2m)}\cos (t))dt +\int_{\pi/2}^{\pi}(e^{-t/(2m)}\cos (t))dt \right)\\ &=e^{-(k\pi)/m} (1-e^{-\pi/(2m)})\left( \int_{0}^{\pi/2}(e^{-t/(2m)}\cos (t))dt -\int_{0}^{\pi/2}(e^{-(\pi-t)/(2m)}\cos (t))dt \right)\\ &=e^{-(k\pi)/m} (1-e^{-\pi/(2m)})\left( \int_{0}^{\pi/2}(e^{-t/(2m)}-e^{-(\pi-t)/(2m)})\cos (t)dt \right)\\ &=e^{-(k\pi)/m} (1-e^{-\pi/(2m)})\left( \int_{0}^{\pi/2}e^{-t/(2m)}(1-e^{-(\pi-2t)/(2m)})\cos (t)dt \right)\\ \end{array} $

Now, at last, everything is positive (P(error) > .25), so we can get bounds and then work our way back.

If we assume that $m$ is large, then, using $e^{-z} \approx 1-z$ for small $z$, making each successive approximation worse, $e^{-t/(2m)}(1-e^{-(\pi-2t)/(2m)}) \approx (1-t/(2m))((\pi-2t)/(2m)) \approx (1-t/(2m)(\pi/(2m)) \approx \pi/(2m) $.

Using this last approximation, $I(m, k) \approx e^{-(k\pi)/m} (1-e^{-\pi/(2m)}) \int_{0}^{\pi/2}(\pi/(2m)\cos (t)dt = e^{-(k\pi)/m} (1-e^{-\pi/(2m)}) (\pi/(2m)) $.

Therefore

$\begin{array}\\ I(2m) &=\frac1{2m}\sum_{k=0}^{m-1}I(m, k)\\ &\approx \frac1{2m}\sum_{k=0}^{m-1} e^{-(k\pi)/m} (1-e^{-\pi/(2m)}) (\pi/(2m))\\ &=(1-e^{-\pi/(2m)}) \frac{\pi}{4m^2}\sum_{k=0}^{m-1} e^{-(k\pi)/m} \\ &=(1-e^{-\pi/(2m)}) \frac{\pi}{4m^2}\frac{1-e^{-\pi}}{1- e^{-\pi/m}} \\ &= \frac{\pi(1-e^{-\pi})}{4m^2} \\ &= \frac{\pi(1-e^{-\pi})}{n^2} \\ \end{array} $

Considering the chances that I made some error in all this are non-zero, I do not know if this is correct, but something like this should be true.

Answer 4

A simpler way.

Since $\cos(z) = \Re(e^{iz})$,

$\begin{array}\\ I(n) &=\int_0^\pi e^{-t}\cos nt dt\\ &=\Re \int_0^\pi e^{-t}e^{int} dt\\ &=\Re \int_0^\pi e^{-t+int} dt\\ &=\Re \frac{e^{t(-1+in)}}{-1+in}\big|_0^\pi\\ \end{array} $.

$\begin{array}\\ \frac{e^{t(-1+in)}}{-1+in} =\frac{e^{-t}+e^{int}}{-1+in}\\ =\frac{e^{-t}+\cos(nt)+i\sin(nt)}{-1+in}\frac{-1-in}{-1-in}\\ =\frac{(e^{-t}+\cos(nt)+i\sin(nt))(-1-in)}{-1+n^2}\\ =\frac{-(e^{-t}+\cos(nt)+i\sin(nt))-in(e^{-t}+\cos(nt)+i\sin(nt))}{-1+n^2}\\ =\frac{-e^{-t}-\cos(nt)-i\sin(nt))-ine^{-t}-in\cos(nt)+n\sin(nt)}{-1+n^2}\\ =\frac{-e^{-t}-\cos(nt)+n\sin(nt)-i(\sin(nt)+ne^{-t}+n\cos(nt))}{-1+n^2}\\ \end{array} $

Therefore

$\begin{array}\\ I(n) &=\frac{-e^{-t}-\cos(nt)+n\sin(nt)}{-1+n^2}\big|_0^\pi\\ &=\frac{(-e^{-\pi}-\cos(n\pi)+n\sin(n\pi))-(-e^{0}-\cos(0)+n\sin(0))}{-1+n^2}\\ &=\frac{(-e^{-\pi}-\cos(n\pi)+n\sin(n\pi))-(-1-1)}{-1+n^2}\\ &=\frac{2+(-e^{-\pi}-\cos(n\pi)+n\sin(n\pi))}{-1+n^2}\\ \end{array} $

If $n$ is even, $I(n) =\frac{2+(-e^{-\pi}-1)}{-1+n^2} =\frac{1-e^{-\pi}}{-1+n^2} $.

If $n$ is odd, $n=2m+1$,

$\begin{array}\\ I(n) &=\frac{2+(-e^{-\pi}-(\cos((2m+1)\pi)+n\sin((2n+1)\pi))}{-1+n^2}\\ &=\frac{2+(-e^{-\pi}-n\sin((2m+1)\pi))}{-1+n^2}\\ &=\frac{2+(-e^{-\pi}-n(-1)^m))}{-1+n^2}\\ &=\frac{2-e^{-\pi}-n(-1)^{\lfloor n/2\rfloor)}}{-1+n^2}\\ \end{array} $.

If $n=4m+1$, $I(n) =\frac{2-e^{-\pi}-n}{-1+n^2} =\frac{2-n-e^{-\pi}}{-1+n^2} $.

If $n=4m+3$, $I(n) =\frac{2-e^{-\pi}+n}{-1+n^2} =\frac{2+n-e^{-\pi}}{-1+n^2} $.