Instead, let’s first evaluate the integral
$\displaystyle I(a)=\int_0^{\infty} \cos (\pi t) e^{-a t} d t \tag*{} $
where $a>0$.
Using Euler identity : $e^{xi}=\cos x+i\sin x$, we have
$\displaystyle \begin{aligned}I(a) & =\int_0^{\infty} \cos (\pi t) e^{-a t} d t \\& =\Re \int_0^{\infty} e^{(\pi i-a) t} d t \\& =\Re\left[\frac{e^{(\pi i-a) t}}{\pi i-a}\right]_0^{\infty} \\& =\Re\left(\frac{1}{a-\pi i}\right) \\& =\frac{a}{a^2+\pi^2}\end{aligned}\tag*{} $
Then differentiating $I(a)$ w.r.t. $a$ yields
$\displaystyle I^{\prime}(a)=-\int_0^{+\infty} t \cos (\pi t) e^{-a t} d t= \frac{a^2-\pi^2}{\left(a^2+\pi^2\right)^2}\tag*{} $
Again, differentiating $I’(a)$ w.r.t. $a$ further gives
$\displaystyle I^{\prime \prime}(a)=\int_0^{+\infty} t^2 \cos (\pi t) e^{-a t} d t= \frac{2 a\left(a^2-3 \pi^2\right)}{\left(a^2+\pi^2\right)^3}\tag*{} $
In general, for any non-negative integer $n$ and $m>0.$
$\displaystyle \boxed{ \left.\int_0^{+\infty} t^n \cos (\pi t) e^{-m t}dt=(-1)^n \frac{d^n}{d a^n}\left(\frac{a}{a^2+\pi^2}\right)\right|_{a=m}}\tag*{} $
My Question
Is there a closed form for
$$\int_0^{+\infty} t^n \cos (\pi t) e^{-m t}dt \textrm{ or } \left.(-1)^n \frac{d^n}{d a^n}\left(\frac{a}{a^2+\pi^2}\right)\right|_{a=m}$$
for any non-negative integer $n$ and $m>0$?
Best Answer
\begin{align} &\int_0^{+\infty} t^n \cos (\pi t) e^{-a t}dt \\ =&\ (-1)^n \frac{d^n}{d a^n}\left(\frac{a}{a^2+\pi^2}\right) = (-1)^n \frac{d^{n+1}}{d a^{n+1}}\Re \ln(a+i\pi)\\ =&\ \Re \frac{n!}{(a+i\pi)^{n+1}} = \frac{n! \cos[(n+1)\tan^{-1}\frac{\pi}{a}]}{(a^2+\pi^2)^\frac{n+1}{2}} \end{align}