<s>[INST] 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?
 [/INST] 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}
$.
</s>