Hey everyone. I thought I'd post this theorem, perhaps someone has some uses for it.

Theorem:

A.) If \( F \) is holomorphic for \( \Re(z) > -a-\epsilon>-1 \) for some \( a> 0 \) and \( F(z) < C e^{\alpha |\Im(z)| + \rho|\Re(z)|} \) for \( 0 \le \alpha < \pi/2 \) and \( \rho \ge 0 \).

B.) for some \( b > 1 \) and \( n \in \mathbb{N} \) we have \( F(\log_b(1+n)) = F(1+n) - 1 \)

Then, for \( \Re(z) > 1-a \) we have \( F(\log_b(1+z)) = F(1+z) - 1 \)

Proof:

Well this is rather easy:

\( \frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)F(1-\xi)x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(1+n)\frac{(-x)^n}{n!} \)

Which follows by cauchy's residue formula and the bounds of F (the gamma function along with x small enough pulls the arc next to our line integral to zero at infinity). For those who don't see,

\( \G(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+z)} + \int_1^\infty e^{-t}t^{z-1}\,dt \)

where the right term is entire in z and only contribute asymptotics, observe stirlings asymptotic formula

\( \G(z) \sim \sqrt{2\pi} z^{z-1/2}e^{-z} \)

Therefore this holds.

Now observe that by a similar argument:

\( \frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)F(\log_b(1-\xi))x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(\log_b(1+n))\frac{(-x)^n}{n!}= \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x) \)

And of course, by another similar argument:

\( \frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)(F(1-\xi)-1)x^{-\xi}\,d\xi = \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x) \)

Therefore since the kernel of this integral transform is zero (its a modified fourier transform). On the line \( [a-i\infty,a+i\infty] \) we have \( F(1-\xi) -1 = F(\log_b(1-\xi)) \). Therefore since both functions are analytic we get the desired. \( \box \)

I'm wondering, does anyone see any uses for this?

I know with some formal manipulation we can say that, if \( G(a_n) = 1+n \) and \( G(a_n-1) = \log_b(n+1) \) and \( G \) is holo and is invertible which satisfies the bounds above. Then \( b^{G(z)} = G(z+1) \)

Theorem:

A.) If \( F \) is holomorphic for \( \Re(z) > -a-\epsilon>-1 \) for some \( a> 0 \) and \( F(z) < C e^{\alpha |\Im(z)| + \rho|\Re(z)|} \) for \( 0 \le \alpha < \pi/2 \) and \( \rho \ge 0 \).

B.) for some \( b > 1 \) and \( n \in \mathbb{N} \) we have \( F(\log_b(1+n)) = F(1+n) - 1 \)

Then, for \( \Re(z) > 1-a \) we have \( F(\log_b(1+z)) = F(1+z) - 1 \)

Proof:

Well this is rather easy:

\( \frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)F(1-\xi)x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(1+n)\frac{(-x)^n}{n!} \)

Which follows by cauchy's residue formula and the bounds of F (the gamma function along with x small enough pulls the arc next to our line integral to zero at infinity). For those who don't see,

\( \G(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+z)} + \int_1^\infty e^{-t}t^{z-1}\,dt \)

where the right term is entire in z and only contribute asymptotics, observe stirlings asymptotic formula

\( \G(z) \sim \sqrt{2\pi} z^{z-1/2}e^{-z} \)

Therefore this holds.

Now observe that by a similar argument:

\( \frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)F(\log_b(1-\xi))x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(\log_b(1+n))\frac{(-x)^n}{n!}= \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x) \)

And of course, by another similar argument:

\( \frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)(F(1-\xi)-1)x^{-\xi}\,d\xi = \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x) \)

Therefore since the kernel of this integral transform is zero (its a modified fourier transform). On the line \( [a-i\infty,a+i\infty] \) we have \( F(1-\xi) -1 = F(\log_b(1-\xi)) \). Therefore since both functions are analytic we get the desired. \( \box \)

I'm wondering, does anyone see any uses for this?

I know with some formal manipulation we can say that, if \( G(a_n) = 1+n \) and \( G(a_n-1) = \log_b(n+1) \) and \( G \) is holo and is invertible which satisfies the bounds above. Then \( b^{G(z)} = G(z+1) \)