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Consider the function $\phi(u) = 2 \sum_{n=1}^\infty (2n^4 \pi^2 e^{9u/2} - 3n^2 \pi e^{5u/2}) e^{-n^2\pi e^{2u}}$.

This appears in Titchmarsh's "The Theory of The Riemann Zeta Function." On page 255, he says $\phi$ is an even function of $u$. However, I do not see how that is true.

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1 Answer 1

This can be treated using Mellin transforms. Introduce $$f(x) = 2 \sum_{n=1}^\infty (2n^4\pi^2 x^{9/2} - 3n^2\pi x^{5/2}) e^{-n^2\pi x^2},$$ so that we seek to prove $f(x) = f(1/x).$ Split this into $$f(x) = f_1(x) - f_2(x)$$ where $$f_1(x) = 4\pi^2 x^{1/2} \sum_{n=1}^\infty (nx)^4 e^{-n^2\pi x^2}$$ and $$f_2(x) = 6\pi x^{1/2} \sum_{n=1}^\infty (nx)^2 e^{-n^2\pi x^2}$$ Now we have $$\mathfrak{M}\left(x^4 e^{-\pi x^2}; s\right) = \frac{1}{2} \frac{\Gamma(s/2+2)}{\pi^{s/2+2}}$$ and $$\mathfrak{M}\left(x^2 e^{-\pi x^2}; s\right) = \frac{1}{2} \frac{\Gamma(s/2+1)}{\pi^{s/2+1}}$$ It follows that $$\mathfrak{M}\left(\sum_{n=1}^\infty (nx)^4 e^{-\pi (nx)^2}; s\right) = \frac{1}{2} \frac{\Gamma(s/2+2)}{\pi^{s/2+2}}\zeta(s)$$ and $$\mathfrak{M}\left(\sum_{n=1}^\infty (nx)^2 e^{-\pi (nx)^2}; s\right) = \frac{1}{2} \frac{\Gamma(s/2+1)}{\pi^{s/2+1}}\zeta(s)$$ and finally $$\mathfrak{M}\left(f_1(x); s\right) = 2\pi^2 \frac{\Gamma(s/2+9/4)}{\pi^{s/2+9/4}}\zeta(s+1/2) = 2\frac{\Gamma(s/2+9/4)}{\pi^{s/2+1/4}}\zeta(s+1/2)$$ and $$\mathfrak{M}\left(f_2(x); s\right) = 3\pi \frac{\Gamma(s/2+5/4)}{\pi^{s/2+5/4}}\zeta(s+1/2) = 3\frac{\Gamma(s/2+5/4)}{\pi^{s/2+1/4}}\zeta(s+1/2).$$ The conclusion is that $$\mathfrak{M}\left(f_1(x)-f_2(x); s\right) = (2(s/2+5/4)-3) \Gamma(s/2+5/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}} \\ = (s-1/2) \Gamma(s/2+5/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}} \\ = (s-1/2)(s/2+1/4) \Gamma(s/2+1/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}} = \frac{1}{2} (s^2-1/4) \Gamma(s/2+1/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}}.$$ Now by Mellin inversion we have that $$f(x) = \frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} \frac{1}{2} (s^2-1/4) \Gamma(s/2+1/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}} \frac{ds}{x^s}.$$ Fortunately the pole of $\zeta(s+1/2)$ at $s=1/2$ gets canceled by the term $s^2-1/4$, so observing exponential decay we may shift this integral to the imaginary axis, getting $$f(x) = \frac{1}{2\pi i} \int_{-i\infty}^{+i\infty} \frac{1}{2} (s^2-1/4) \Gamma(s/2+1/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}} \frac{ds}{x^s}.$$ Now from the functional equation of the Riemann zeta function we have that (substitute $s+1/2$ for $s$) $$ \frac{1}{2} (s^2-1/4) \Gamma(s/2+1/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}} = \frac{1}{2} (s^2-1/4) \Gamma(1/4-s/2) \frac{\zeta(1/2-s)}{\pi^{1/4-s/2}}.$$ But $$f(1/x) = \frac{1}{2\pi i} \int_{-i\infty}^{+i\infty} \frac{1}{2} (s^2-1/4) \Gamma(s/2+1/4) \frac{\zeta(s+1/2)}{\pi^{s/2+1/4}} x^s ds.$$ Put $s=-t$ in this integral to get $$ f(1/x) = - \frac{1}{2\pi i} \int_{+i\infty}^{-i\infty} \frac{1}{2} (t^2-1/4) \Gamma(-t/2+1/4) \frac{\zeta(-t+1/2)}{\pi^{-t/2+1/4}} x^{-t} dt.$$ Substituting from the functional equation and arranging signs, we see that indeed $f(x)=f(1/x),$ as claimed. QED.

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Presumably Titchmarsh did not do this whole calculation but had some other criterion that allowed him to spot eveness instantly. Wouldn't we all love to know what that was. Fascinating how a somewhat irregular-looking sum turns out to encapsulate the functional equation of the Riemann Zeta Function. –  Marko Riedel Jun 13 '13 at 22:39
Wow, thanks! Could you explain how you were able to come up with all of that? –  Alan C Jun 25 '13 at 11:58
@AlanC The key is to be able to recognize harmonic sums, once you have that, the rest is fairly mechanical. –  Marko Riedel Jun 25 '13 at 16:55
Sorry, what do you mean by "harmonic sum"? A sum that arises when dealing with Fourier series? –  Alan C Jun 26 '13 at 0:10
@AlanC The concept is from page 46 of the seminal "The Average Case Analysis of Algorithms: Mellin Transform Asymptotics" by Flajolet and Sedgewick (INRIA Technical Report 2956.) –  Marko Riedel Jun 26 '13 at 0:35

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