Can we show that the following function is integrable
\begin{align} f(t) =\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{\Gamma \left( \frac{it+1}{1.5} \right) }{\Gamma \left(\frac{ it+1}{2} \right)}, \end{align} where $t \in \mathbb{R}$ and $i=\sqrt{-1}$.
That is can we show that \begin{align} \int_{-\infty}^{\infty} |f(t)| dt<\infty. \end{align}
I was wondering if Stirling's approximation can be used, since this is a complex case?
Note if Stirling's approximation can be used than \begin{align} f(t) \approx \sqrt{\frac{1.5}{2}}\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{ \left( \frac{it+1}{1.5 e} \right)^{\frac{1+it}{1.5}} }{ \left(\frac{ it+1}{2 e} \right)^{\frac{1+it}{2}}}. \end{align}
Note that \begin{align} |f(t)| &\approx \left| \frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}}\right| \left| \frac{ \left( \frac{it+1}{1.5 e} \right)^{\frac{1+it}{ 1.5}} }{ \left(\frac{ it+1}{2 e} \right)^{\frac{1+it}{2}}}\right|\\ &=\left| \frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}}\right| \left| \frac{ \left( it+1\right)^\frac{1+it}{1.5} }{ \left( it+1 \right)^{\frac{1+it}{2}}}\right| \left|\frac{ \left(2 e\right)^{\frac{1+it}{2 }}} {(1.5 e)^{\frac{1+it}{ 1.5}}} \right| \end{align}
Also we have that \begin{align} &\left|\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}}\right|= 2^{\frac{2}{3}-\frac{1}{2}} \\ & \left|\frac{ \left(2 e\right)^{\frac{1+it}{2 }}} {(1.5 e)^{\frac{1+it}{ 1.5}}} \right| =\frac{ \left(2 e\right)^{\frac{1}{2 }}} {(1.5 e)^{\frac{1}{ 1.5}}} \end{align}
So, in the end we if everthing is corect we have to show that \begin{align} g(t)=\left| \frac{ \left( it+1\right)^\frac{1+it}{1.5} }{ \left( it+1 \right)^{\frac{1+it}{2}}}\right| \end{align} is integrable, but I am not sure how to show if the above equation is integrable or not?
Any ideas would be greatly appreciated. Thank you.
Edit The integrability of $g(t)$ has been shown in one of the answers.
My question now is the following: Since we have that \begin{align} |f(t)| =|g(t)| +e \end{align} and $g(t)$ is integrable does this mean that $f(t)$ is inegrable? Can we prove that the error term $e$ is also integrable?
We have $$\left| \frac{ \left( it+1\right)^\frac{1+it}{1.5} }{ \left( it+1 \right)^{\frac{1+it}{2}}}\right| = |(1+it)^{\frac{1+it}{6}}|= \left|\exp\left(\left(\frac{1+it}{6}\right)\ln(1+it)\right)\right|.$$ Remember that $|e^z|=e^{\Re(z)}$ and $$\Re\left(\left(\frac{1+it}{6}\right)\ln(1+it)\right)=\frac{\ln(\sqrt{1+t^2})}{6}-\frac{t\arg(1+it)}{6}.$$ (Here we have used the definition of the complex logarithm $\ln(z) = \ln|z|+i\arg(z)$.) Hence $$g(t) = (1+t^2)^{1/12} e^{-\arg(1+it)t/6}.$$ If $t>0$ is big enough, we can assume that $\arg(1+it)>\pi/4$, hence $$g(t) < (1+t^2)^{1/12} e^{-\frac{\pi t}{24}}$$ is clearly integrable at $+\infty.$ You can procede in a same way for $-\infty.$