1
$\begingroup$

It appears from the figure 11.3.4 of dlmf that the Struve function $M_1(x)$ is monotonically decreasing for positive $x$. The asymptotic expansion (11.6.2) shows that the limit is $-2/\pi$. So it seems clear that

$$M_1(x)>-\frac{2}{\pi},\qquad x>0$$

But, how to prove this? Any hints or references much appreciated.

Edit: As indicated above, I am using dlmf definitions (11.2.2, 11.2.6):

$$M_\nu(z)=L_\nu(z)-I_\nu(z)$$ where $I$ is the modified Bessel function and

$$ L_\nu(z)=\left(\frac{z}{2}\right)^{\nu+1}\sum_{n=0}^\infty\frac{\left(\frac{z}{2}\right)^{2n}}{\Gamma\left(n+\frac32\right)\Gamma\left(n+\nu+\frac32\right)}$$

is the modified Struve function. The asymptotic expansion is given as (11.6.2):

$$ M_\nu(z)\sim\frac{1}{\pi}\sum_{k=0}^\infty(-1)^{k+1}\frac{\Gamma\left(k+\frac12\right)\left(\frac{z}{2}\right)^{\nu-2k-1}}{\Gamma\left(\nu+\frac12-k\right)},\qquad |z|\to\infty,\quad|\mbox{ph}(z)|\leq\frac{\pi}{2}-\delta$$

Improve this question
$\endgroup$
5
  • $\begingroup$ Is your $M$ written as $K$ in the reference? $\endgroup$
    – Fan Zheng
    Commented Aug 2, 2017 at 10:03
  • $\begingroup$ No, $M$ is $M$. Why do you think I mean $K$? $\endgroup$
    – Carl
    Commented Aug 2, 2017 at 10:47
  • $\begingroup$ Following @FanZheng question, in both Wiki and Wolfram there are many different definitions of Struve and modified Struve functions. I think it'd be best if you write the definition of $M_1$ in the oringinal post. $\endgroup$
    – Amir Sagiv
    Commented Aug 2, 2017 at 10:48
  • $\begingroup$ OK, but it should be clear I am using dlmf definitions. $\endgroup$
    – Carl
    Commented Aug 2, 2017 at 10:48
  • $\begingroup$ @AmirSagiv Actually I checked wiki and wolfram and it seems both have the same definition of the H and L functions as dlmf. $\endgroup$
    – Carl
    Commented Aug 3, 2017 at 8:15

0

Reset to default

You must log in to answer this question.