# Integrability of $f(t) =\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{\Gamma \left( \frac{it+1}{1.5} \right) }{\Gamma \frac{ it+1}{2} }$

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?

• @polfosol Yes, $i$ here is the imaginary number. I will add this. – Boby Jan 5 '17 at 16:25
• Why the title is different than the question? – polfosol Jan 5 '17 at 16:31
• @polfosol Yes, you right. Corrected that too. – Boby Jan 5 '17 at 16:41
• "Experimentally" (i.e. by plotting the function), it appears that $\ln |f(t)| \approx -|t|/4$ asymptotically; if this is actually true, it should be integrable. No idea how to prove it, though. – Michael Seifert Jan 5 '17 at 22:17
• @MichaelSeifert Thanks. I also think it should be integrable. – Boby Jan 5 '17 at 22:50

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 $-\infty.$

• Can you give more explanation on how you deal with $e^{-\frac{t}{6} {\rm arg}(1+it)}$? – Boby Jan 6 '17 at 22:01
• I made a small mistake, now it should be clear. – C. Dubussy Jan 6 '17 at 22:18
• Question: Does strings approximation for gamma function hold for complex domain? – Boby Jan 6 '17 at 22:48
• Yes, if $\Re(z)>0$, you can see here for ex. en.wikipedia.org/wiki/Stirling's_approximation – C. Dubussy Jan 6 '17 at 23:43
• So, you think integrability is not effect by Striling's approximation? In a sense that the error also integrable? – Boby Jan 6 '17 at 23:47