# Fourier Analysis – Asymptotic \$ sum_ {n geq 1} frac { sin n ^ 2 t} {n ^ 2} \$

This is the function of Riemann.

I would like to determine your asymptotes as $$t rightarrow 0 ^ +$$.

First leave $$t = x ^ 2$$, so that we can treat the series.

$$sum_ {n geq 1} frac { sin n ^ 2 x ^ 2} {n ^ 2}.$$

We use the Mellin transform. For that, we need Mellin to transform the function. $$without x ^ 2$$. By changing variables, and using the fact that the Mellin transform of the function $$without x$$ is $$Gamma (s) without frac { pi s} {2}$$, we obtain

$$mathcal {M}[sin x^2 ;s ] = frac {1} {2} Gamma ( frac {s} {2}) without frac { pi s} {4}, ; ; ; 0

The Mellin transform of the original series is then.

$$frac {1} {2} Gamma ( frac {s} {2}) sin frac { pi s} {4} zeta (s + 2), ; ; ; 0

Now we use the investment formula to recover the original series. Is

$$frac {1} {2 pi i} int_C x ^ {- s} frac {1} {2} Gamma ( frac {s} {2}) sin frac { pi s} {4} zeta (s + 2) ds,$$

Where can we take the integration outline. $$C$$ to be the vertical line from $$1-i infty$$ to $$1 + i infty$$.

The common trick is then to change the integration line to the left. In this process, the integrand poles are rounded one by one.

Now, the problem is that the gamma function has poles in $$s = 0, -2, -4, -6,$$ etc. The zeta function has only one pole in $$s = -1$$. The sine function has zeros in $$s = 0, -4, -8,$$ etc. The zeta function has zeros in $$s = -4, -6, -8,$$ etc.

After cancellation, the remaining poles are $$s = 0, -1, -2$$. We obtain a quadratic polynomial of $$x$$.

This is obviously wrong. The Riemann function is not differentiable almost everywhere.

Could someone show me where I made a mistake?