Showing $\int_0^\infty t^{z-1} e^{-t}\cos(t)dt=\Gamma(z) 2^{\large \frac{-z}2}\cos\left(\frac{\pi z}{4}\right)$

Question

Showing that $$\int_0^\infty t^{z-1} e^{-t}\cos(t)dt=\Gamma(z) 2^{\large \frac{-z}2}\cos\left(\frac{\pi z}{4}\right)$$ where $\displaystyle z \in \mathbb C$ and $\Re(z)>0$

My incomplete attempt

According to $\displaystyle \cos(t)= \sum_{k=0}^{\infty} \frac{(-1)^n}{(2n)!}t^{2n} \quad( t \in \mathbb R )$, we can write $\displaystyle \begin{align*} \int_0^\infty t^{z-1} e^{-t}\cos(t)dt &=\int_0^\infty t^{z-1} e^{-t} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}t^{2n} dt\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \int_0^\infty t^{z+2n-1} e^{-t} dt\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \Gamma(z+2n)\\ &=\Gamma(z) \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \frac{\Gamma(z+2n)}{\Gamma(z)}\\ &=\Gamma(z) \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} z^{(2n)} \qquad because \quad \frac{\Gamma(z+n)}{\Gamma(z)}=z^{(n)}=z(z+1)...(z+n-1)\\ &=\Gamma(z) \sum_{n=0}^{\infty} \binom{-z}{2n} (-1)^n \qquad because \quad z^{(p)}=(-1)^p\binom{-z}{p}\\ &=\ ...\\ \end{align*} $ But I can't make the $cos$ appear.

Answer 1

Using $$e^{-(a + i \, b) \, t} = e^{-a t} \, (\cos(b t) + i \, \sin(b t))$$ and $$\int_{0}^{\infty} e^{- x \, t} \, t^{s-1} \, dt = \frac{\Gamma(s)}{x^{s}}$$ then \begin{align} \int_{0}^{\infty} e^{-(a+i b) t} \, t^{x-1} \, dt &= \frac{\Gamma(x)}{(a + i b)^{x}} = \frac{\Gamma(x)}{(a^{2} + b^{2})^{x}} \, (a - i b)^{x} \\ &= \frac{\Gamma(x)}{(a^{2} + b^{2})^{x/2}} \, e^{- x \, \tan^{-1}(b/a)} \\ &= \frac{\Gamma(x)}{(a^{2} + b^{2})^{x/2}} \, \left( \cos\left( x \, \tan^{-1}(b/a)\right) - i \sin\left(x \tan^{-1}(b/a)\right) \right). \end{align} From this: \begin{align} \int_{0}^{\infty} t^{x-1} \, e^{-a t} \, \cos(b t) \, dt &= \frac{\Gamma(x)}{(a^{2} + b^{2})^{x/2}} \, \cos\left( x \, \tan^{-1}\left(\frac{b}{a}\right)\right) \\ \int_{0}^{\infty} t^{x-1} \, e^{-a t} \, \sin(b t) \, dt &= \frac{\Gamma(x)}{(a^{2} + b^{2})^{x/2}} \, \sin\left( x \, \tan^{-1}\left(\frac{b}{a}\right)\right). \end{align} Setting $a$ and $b$ to one leads to the desired result.

Answer 2

Your integral is equal to \begin{align} I=\Re\int_{0}^{\infty} t^{z-1}e^{-t(1-i)}\,dt \end{align} Ramanujan's master theorem states that if $f(x)$ can be expanded in a power series in this form, \begin{align} f(x)=\sum_{n=0}^{\infty} \frac{\phi(n)}{n!}(-x)^n \end{align} Then the Mellin transform is given by $$\int_0^{\infty} x^{z-1}f(x)\,dx=\Gamma(z)\phi(-z) $$ Can you take it on from here? Hint: Remember the exponential Maclaurin series: $$e^s=\sum_{n=0}^{\infty} \frac{s^n}{n!}\qquad s\in\mathbb{C}$$