Estimate from below $\int_0^\pi e^{-t}\cos nt dt$ without calculate it. 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?
 A: 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.
A: $$\begin{eqnarray*} I &=& \frac{1}{n}\int_{0}^{n\pi}e^{-t/n}\cos t\,dt =\frac{1}{n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}e^{-t/n}\cos t\,dt\\&=&\frac{1}{n}\int_{0}^{\pi}\cos t\,e^{-t/n}\sum_{j=0}^{n-1}(-1)^j e^{-j\pi/n}\,dt\tag{1}\end{eqnarray*}$$
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 $\Omega\left(\frac{1}{n^2}\right)$.
A: 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.
A: 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}
$.
