Integration of Power Series. When is the Function in $L_1$?

Suppose we are given a function $f$ in terms of its power series \begin{align} f(x)= \sum_{n+0}^\infty a_n x^n. \end{align} We assume that we know all $a_n$'s and $f$ has infinite radius of convergense.

Can we say based on $a_n$ if the function is integrable or not? That is if \begin{align} \int_{\mathbb{R}} |f(x)| dx<\infty. \end{align}

For example, if $a_n \ge 0$ then we have that \begin{align} f(x) \ge a_0+a_1 x,\ x >0 , \end{align} and the function is not integrable.

The case that I am interested is when $a_n$'s have alternating sign. Specifically, can we determine if the following $f(x)$ is integrable \begin{align} f(x)= \sum_{k=0}^\infty \frac{\cos( \frac{\pi}{2}k)\ (k+1)^{\frac{k+1}{6}}}{k!} x^k. \end{align}

Here is the plot of $f(x)$ where the series was computed up to $N=600$.

where blue curve is $f(x)$ and red curve is $sinc(\pi/2 x)= \frac{\sin(\pi/2 x)}{\pi/2 x}$.

It seems that $f(x)$ has a tail that decays faster than $sinc(\pi/2 x)$.

Moreover, next we give a plot of $f(x) \cdot x$ and $f(x) \cdot x^2$

where blue curve is $f(x) \cdot x$ and red curve is $f(x) \cdot x^2$.

This seems to indicate that $f(x)$ decreases faster than $\frac{1}{x^2}$ which would imply that $f(x)$ is integrable.

Therefore, it would also be interesting to show that \begin{align} \lim_{x \to \infty} f(x)=0, \end{align} which at this moment I do not know how to do.

Thanks you.

**Edit: ** Please see a possible solution via Mellin transform.

• If the power series converges in all of $\mathbb{R}$, the function is analytic, so it should automatically be integrable, unless I'm being really stupid, which I conceivably might be. – Daminark Dec 26 '16 at 4:26
• If $a_n > 0$ for even $n$, $f(x) + f(-x) = 2 \sum_{n\ \text{even}} a_n x^n \ge a_2 x^2$ for $x > 0$, and $f(x)$ and $f(-x)$ can't both be integrable on $[0,\infty)$, and $f$ can't be integrable on $\mathbb R$. – Robert Israel Dec 26 '16 at 5:33
• sinc is not integrable, although its improper integral converges. For integrability you'll want a decay better than the $1/|x|$ of sinc, maybe $1/|x|^p$ for some $p > 1$. The picture might suggest that this is the case, but of course it's not a proof. – Robert Israel Jan 3 '17 at 20:49
• If that's true (which I'm not convinced of), then it is integrable. – Robert Israel Jan 3 '17 at 23:45
• @boby Yeah sure, I don't have time at the moment, but I'll make it an answer and add some references and fill in a few details later. – user335907 Jan 4 '17 at 21:18

To start consider mellin's inversion theorem.

Mellin's inversion theorem can be stated as, if

$$f(x) = \frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma + i\infty}F(z)x^{-z}\,dz$$

Assuming everything is convergent, $0 < \sigma < 1$, $F$ holomorphic, then

$$F(z) = \int_0^\infty f(x)x^{z-1}\,dx$$

for $0 < \Re(z) < 1$

Ramanujan's master theorem gives us a really beautiful result based off this. If $z\Gamma(-z)g(z) \to 0$ as $z \to \infty$ when $|\arg(z)| < \pi/2$ and $g(z) = O(e^{\tau|\Im(z)|})$ for $\tau < \pi/2$, then

Ramanujan's Master theorem says

$$f(x) = \sum_{n=0}^\infty g(n)\frac{(-x)^n}{n!} = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma+ i \infty} \Gamma(z)g(-z)x^{-z}\,dz$$

If you need more justification of this fact look at Ramanujan's master theorem, even wikipedia has something on this. There are a few ways of stating this, but the version I'm using is perfectly valid.

What is great is that

$$\int_0^\infty |f(x)| x^{-\sigma}\,dx < \infty$$

Which is another consequence of Mellin's inversion theorem.

Now your function $F(z) = \cos(\pi/2 z)(z+1)^{(z+1)/6}$ satisfies these bounds. It is rather trivial to show this, if need be I'll edit this in. By the above

$$f(x) = \sum_{n=0}^\infty \cos(\pi/2n)(n+1)^{(n+1)/6}\frac{(-x)^n}{n!}$$

satisfies for $0 < \Re(z) < 1$

$$\int_0^\infty f(x)x^{z-1}\,dx = \Gamma(z)F(z)$$

Therefore off the start, your function $f$ is in a weighted $L^{1}$ space (with weight $x^{-\sigma}$ (this is also part of Mellin's inversion theorem). All the odd powers disappear, meaning the function is symmetric about zero. Therefore

$$\int_{-\infty}^\infty |f(x)|x^{-\sigma}\,dx < \infty$$

for all $0 < \sigma < 1$. I'll leave it to you to show the subtleties required in making it $L^1$. If you can't, I'll show it .

• I just wanted to break up your argument into logical steps. Let me know if the following is correct. The steps are: 1) Identify that $g(z)=\cos(\pi/2 z) (z+1)^{(z+1)/6}$ has the property that $z \Gamma(-z) g(z) \to 0$ as $z \to \infty$. 2) By Ramanujan's master theorem this implies that the function $f(z)= \sum_{n=0}^\infty g(n) \frac{(-x)^n}{n!}$ can be expressed as $f(z)= \frac{1}{2\pi i} \int_{\sigma-i \infty}^{\sigma+i \infty} \Gamma(z) g(-z) x^{-z} dz$. 3) This is the step I am confused about. How do we now know that $\int_{-\infty}^\infty |f(x)| x^{-\sigma} dx<\infty$??? – Boby Jan 5 '17 at 18:54
• @Boby Am I being stupid or is this just because $f(z)$ is the Mellin transform of $\Gamma(z)g(-z)$, so $\int_0^\infty f(x)x^{z-1}\,dx = \Gamma(z)F(z)$? – Anon Jan 6 '17 at 2:10
• @Anon I am not sure. I guess the claims is that if $f(z)$ is Mellin transform of $\Gamma(z)g(-z)$ then $f(z)$ satisfies $\int_{0}^\infty |f(z)| x^{-\sigma} dx<\infty$. But I don't know where is this comming from? – Boby Jan 6 '17 at 3:47
• @Boby I think it's from Mellin's inversion theorem which he mentioned. I think that will give you his second-last formula (by applying the inversion theorem to his third formula which has the form of an inverse Mellin transform), which (with what he says at the beginning) I think gives the last formula. – Anon Jan 6 '17 at 4:18
• Yeah, it's all because of Mellin's inversion theorem. $\int_0^\infty f(x)x^{z-1}\,dx = \Gamma(z)F(z)$, which isn't trvial to prove. If I showed the whole thing it would take like 5 pages. Just look up Mellin's inversion theorem, and Ramanujan's master theorem. Both of these results should clear up your confusion. – user335907 Jan 6 '17 at 18:32