Showing $\int_0^\infty t^{z-1} e^{-t}\cos(t)dt=\Gamma(z) 2^{\large \frac{-z}2}\cos\left(\frac{\pi z}{4}\right)$ 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.
 A: 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:
$$e^s=\sum_{n=0}^{\infty} \frac{s^n}{n!}\qquad s\in\mathbb{C}$$
A: 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.
