View source for H (nu)(x) geq 0 for x gt 0 and nu geq 1/2

==Theorem==
The following formula holds for $x>0$ and $\nu \geq \dfrac{1}{2}$:
$$\mathbf{H}_{\nu}(x) \geq 0,$$
where $\mathbf{H}$ denotes a [[Struve function]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=D/dz(z^(-nu)H (nu))=1/(sqrt(pi)2^(nu)Gamma(nu+3/2))-z^(-nu)H (nu+1)|next=H_(-(n+1/2))(z)=(-1)^n J_(n+1/2)(z) for integer n geq 0}}: $12.1.14$

[[Category:Theorem]]
[[Category:Unproven]]