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Can we determine the sign of the following function \begin{align} f(a,k)=\int_{-\infty}^\infty \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{\frac{it}{3/2}} \Gamma \left( \frac{it +1}{3/2}\right) } \frac{1}{(a+it)^k} dt, \end{align} where $a\neq 0$ and $k \ge 1$ is some positive integer.

The conjecture is that the sign of the integral is equal to \begin{align} {\rm sign } (f(a,k))={\rm sign}(a)^k. \end{align}

Perhpas the following limit can be usefull. By using a method in this question it is not difficult to see that \begin{align} \left | \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{\frac{it}{3/2}} \Gamma \left( \frac{it +1}{3/2}\right) } \right| \to O( e^{- (\frac{3}{2}-\frac{2}{3}) t}) \text{ as } t \to \infty. \end{align}

Thanks

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  • $\begingroup$ $1.5 = 3/2$, so why haven't you cancelled those powers of $2$? $\endgroup$ – Robert Israel Feb 20 '17 at 19:43
  • $\begingroup$ @RobertIsrael Sorry, there is a typo. It should be $2/3$ instead of $3/2$ $\endgroup$ – Lisa Feb 20 '17 at 19:46
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    $\begingroup$ Numerical plotting seems to indicate that the first part of the integrand decays rapidly outside the interval $[-5,5]$ or so, and of course $(a+it)^{-k}$ also decreases in magnitude away from $0$, so numerical methods should yield a sufficiently accurate estimate. It appears that the sign is the same as that of $a^k$. $\endgroup$ – Robert Israel Feb 20 '17 at 20:05
  • $\begingroup$ @RobertIsrael I also had a conjecture that the sign is ${\rm sign}(a)^{k}$. But not sure what method can show this. $\endgroup$ – Lisa Feb 20 '17 at 20:18
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Not quite an answer, but a good start. Let's look at

$$f(a,k)=\int_{-\infty}^\infty \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{\frac{it}{3/2}} \Gamma \left( \frac{it +1}{3/2}\right) } \frac{1}{(a+it)^k} dt$$

which can be rewritten as

$$f(a,k)= (-i) \int_{-i\infty}^{i\infty} 2 ^{\frac{5z}{6}}\frac{ \Gamma \left( \frac{3(z +1)}{2}\right) }{ \Gamma \left( \frac{2(z +1)}{3}\right) } \frac{1}{(a+z)^k} dz$$

Observe that $\Gamma \left( \frac{3(z +1)}{2}\right)$ has poles in the left half plane when $\frac{3(z+1)}{2} = n$ and $ n = 0, -1, -2, -3,....$, so when $z = \frac{2}{3}n - 1$ we have a pole. The principal part is

$$\frac{(-1)^n}{n!(\frac{3(z+1)}{2} +n)} = \frac{2}{3}\frac{(-1)^n}{n!(z + 1 + \frac{2}{3}n)}$$

Take a semicircle contour that grows in the left half plane. A simple exercise in Mellin transforms gives that, if $a < 0$

$$f(a,k) = \frac{4\pi}{3}\sum_{n=0}^\infty \frac{(-1)^n2^{-\frac{5}{6}(1+\frac{2}{3}n)}}{n!\Gamma(-\frac{4}{9}n)(a-1-\frac{2}{3}n)^k}$$

Now showing that $\text{sign}(f(a,k)) = \text{sign}(a)^k$ involves talking about this series. Note that some of the terms disappear if $n = 0 \,\mod 9$ (because the Gamma function on the bottom vanishes there). Not sure how you would really approach this, but probably discussing that this series oscillates wildly has something to do with it. It seems obvious though that if $a<0$ then $\text{sign}(f(a,k))^k = \text{sign}(a)^k$.

EDIT: If $a > 0$ there's another term added because $\frac{1}{(a+z)^k}$ has a pole in the left half plane. This needs to be handled in cases, because when $a = -\frac{2}{3}n -1$ the residues get all wonky. I'll leave it to you to find that extra term, which isn't too hard to get at. It just involves taking the $k$'th derivative of the rest of the integrand (something I'm not in the mood to do).

PS: I may have screwed up some arithmetic; the amount of fractions I just crossed out and rearranged in my head had me blurry eyed.

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  • $\begingroup$ Thank you. I will try to figure out what the expression is for $a>0$. When you say that if $a<0$ then $sign f(a,k)=sing(a)^k$ can you tell me how you see it from the series? $\endgroup$ – Lisa Feb 23 '17 at 1:05
  • $\begingroup$ It's just a hunch, but $\frac{(-1)^n}{\Gamma(-\frac{4}{9}n)}$ appears to usually be positive. So that most of the terms have the same sign as $(a-1-\frac{2}{3}n)^k$. The odd terms that aren't positive though shouldn't contribute enough to force the entire sum to be the opposite sign. I'm not sure how you'd phrase that mathematically; it just seems to be the case. I'll try and think about it some more. $\endgroup$ – user335907 Feb 23 '17 at 2:43
  • $\begingroup$ thank you very much for you help. $\endgroup$ – Lisa Feb 23 '17 at 12:31
  • $\begingroup$ Dear James, I posted this question. I was wondering if you could take a look: math.stackexchange.com/questions/2179266/… $\endgroup$ – Lisa Mar 9 '17 at 15:46

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