Is there a closed form of $\int_0^{+\infty} t^n \cos (\pi t) e^{-m t}dt$ for any non-negative integer $n$ and $m>0$?

Question

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 $m>0$?

Answer 1

\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}

Answer 2

Let us define $$I_n(a) = \int_0^\infty t^n e^{-(a+i\pi)t} ~ dt$$ so that your integral equals the real part of $I_m(a)$. By integration by parts, for $n>0$, we get that $$I_n(a) = \frac{n}{a+i\pi}\int_0^\infty t^{n-1}e^{-(a+i\pi)t} ~ dt$$ which, by induction, gives us $$I_n(a) = \frac{n!}{(a+i\pi)^n}I_0(a) = \frac{n!}{(a+i\pi)^{n+1}}$$ We wish to extract the real part of this expression, and writing the denominator in polar coordinates shall help us with this. We have that $$a+i\pi = \sqrt{a^2+\pi^2}\exp\Big(i ~ \arctan\frac{\pi}{a}\Big)$$ hence $$(a+i\pi)^{n+1} = (a^2+\pi^2)^\tfrac{n+1}{2}\exp\Big((n+1)i ~ \arctan\frac{\pi}{a}\Big)$$ and therefore we have $$\color{green}{I_n(a) = \frac{n!}{(a^2+\pi^2)^\tfrac{n+1}{2}}\cos\Big((n+1)\arctan\frac{\pi}{a}}\Big)$$ in agreement with the answer provided by Mathematica in the comments!

Answer 3

As @ MathSensei suggested, I can now write a closed form for the integral as $$ \boxed{\int_0^{+\infty} t^n \cos (\pi t) e^{-at}dt =I^{(n)}(a)=\frac{ n !}{\left(a^2+\pi^2\right)^{\frac{n+1}{2}}}\left[\cos (n+1) \tan ^{-1}\left(\frac{\pi}{a}\right)\right]} $$ by differentiating $I(a)=\Re\left(\frac{1}{a-\pi i}\right)$ w.r.t. $a$ by $n$ times as $$ \begin{aligned} I^{(n)}(a) & =\Re\left[\frac{d^n}{d a^n}\left(\frac{1}{a-\pi i}\right)\right] =\Re\left[\frac{(-1)^n n!}{(a-\pi i)^{n+1}}\right] \end{aligned} $$ Expressing the last term in terms of polar form yields $$ \begin{aligned} (a-\pi i)^{n+1} & =\left[\sqrt{a^2+\pi^2} e^{-i \tan ^{-1}\left(\frac{\pi}{a}\right)}\right]^{n+1} \\ & =\left(a^2+\pi^2\right)^{\frac{n+1}{2}} e^{-i(n+1) \tan ^{-1}\left(\frac{\pi}{a}\right)} \end{aligned} $$ and complete the proof that $$ \int_0^{+\infty} t^n \cos (\pi t) e^{-a t} d t=(-1)^n \frac{d^n}{d a^n} \Re\left(\frac{1}{a-\pi i}\right)= \frac{ n !}{\left(a^2+\pi^2\right)^{\frac{n+1}{2}}}\left[\cos (n+1) \tan ^{-1}\left(\frac{\pi}{a}\right)\right] $$

Answer 4

Franklin's recursive method can actually be extended to non integer values. In fact, on page 498 of the 11th edition of Table of Intgrals, Series, and Products there appears the very general integral (identity 4) $$\int_0^z x^{a-1}\mathrm e^{-bx}\cos(cx)\mathrm dx=\frac{1}{2}(b+\mathrm ic)^{-a}~\gamma\big(a,(b+\mathrm ic)z\big)+\frac{1}{2}(b-\mathrm ic)^{-a}~\gamma\big(a,(b-\mathrm ic)z\big)$$

Where $\gamma(\cdot,\cdot)$ is a lower incomplete Gamma function.

Taking the special case $z\to\infty$, (identity 6) $a=p+1$, $b=q$, $c=\pi$, we get $$\phi(p,q)=\int_0^\infty t^{p} \cos(\pi t)\mathrm e^{-qt}\mathrm dt \\ =\boxed{\frac{\Gamma(p+1)}{(q^2+\pi^2)^{\frac{p+1}{2}}}\cos\left((p+1)\arctan\frac{\pi}{q}\right)}$$

Answer 5

Here is a solution using Cauchy's integral theorem that slightly generalizes some of the other answers: $$\boxed{\int_0^{\infty}t^{\alpha-1}\cos(\pi t)e^{-\beta t}\,dt = \frac{\Gamma(\alpha)}{(\beta^2 + \pi^2)^{\frac{\alpha}{2}}}\cos\left(\alpha\tan^{-1}\left(\frac{\pi}{\beta}\right)\right), \quad\alpha, \beta > 0.}$$

Let $s = \sqrt{\beta^2+\pi^2}$ and $f(z) = z^{\alpha-1}e^{-sz}$ for $z\in \mathbb C$. Further, let $C$ be the union of the four counterclockwise paths $$C_1 := \{xe^{i\theta} :x\in [\varepsilon,R]\}, \quad C_2 := \{Re^{ix}:x\in[0,\theta)\}, \quad C_3 := [\varepsilon,R], \quad C_4:=\{\varepsilon e^{ix} : \quad x\in [0,\theta)\}$$ where $\theta = \tan^{-1}(\pi/\beta)$ and $\varepsilon, R > 0$. Then by Cauchy's theorem, $\int_C f(z)\,dz = 0$ for every $\varepsilon, R > 0$, hence taking $\varepsilon \to 0^+$ and $R \to \infty$ yields $$e^{i\alpha\theta}\int_0^{\infty}t^{\alpha-1}e^{(\beta+i\pi)t}\,dt = \int_0^{\infty}t^{\alpha-1}e^{-st}\,dz = \frac{\Gamma(\alpha)}{s^{\alpha}},\quad (*)$$ where we used the facts $$\lim_{R\to \infty}\left|\int_{C_2} f(z)\,dz\right| \leq \lim_{R\to \infty}R^{\alpha}\int_0^{\theta}e^{-sR\cos x}\,dx = 0,$$ $$\lim_{\varepsilon\to 0^+}\left|\int_{C_4} f(z)\,dz\right| \leq \lim_{\varepsilon\to 0^+}\varepsilon^{\alpha}\int_0^{\theta}e^{-s\varepsilon\cos x}\,dx = 0.$$ Taking the real parts in ($*$) yields the desired result.