# How to show that this function (related to the zeta function) is even?

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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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.