Is it known, and if yes how does one show, that the function

$$ \psi(n):=n\int_0^{+\infty} e^{-x}I_0\left(\frac{x}{n}\right)^{n-1}I_1\left(\frac{x}{n}\right)\mathrm{d}x$$

is decreasing for all $n\ge 3$? Here $I_0$ and $I_1$ denote the modified Bessel functions of the first kind of order $0$ and $1$ respectively.

  • $\begingroup$ May be it is true without integral? Try derivative? $\endgroup$
    – Sergei
    May 30, 2016 at 15:04
  • $\begingroup$ it is a derivative under the integral sigh? $\endgroup$
    – Sergei
    May 31, 2016 at 13:57
  • $\begingroup$ @Sergei: my apologies, you are right concerning the sign of the derivative. It is indeed negative. But how can one show it (without using Mathematica or any computer aid)? The derivative function is rather ugly. $\endgroup$
    – ARC
    May 31, 2016 at 16:13

1 Answer 1


As $I_0^\prime = I_1$, we have $ I_0\left(\frac{x}{n}\right)^{n-1}I_1\left(\frac{x}{n}\right) = \frac{d}{dx} \left(I_0(\frac{x}{n})^n\right)$ , so $$ \psi(n) = - n + n\int_0^\infty I_0\left(\frac{x}{n}\right)^n\exp(-x)dx,\\ = -n +n^2 \int_0^\infty (I_0(y) \exp(-y))^n dy $$
and as $0\leq \exp(-x)I_0(x)\leq1$, expanding $(n+1)^2$, $$ \psi(n+1)-\psi(n) = - 1 + (n+1)^2 \int_0^\infty (I_0(y) \exp(-y))^{n+1} dy \\ -(n)^2 \int_0^\infty (I_0(y) \exp(-y))^{n} dy \\ \leq (2n+1)\int_0^\infty (I_0(y) \exp(-y))^{n+1} dy-1. $$ Now let us use a rough upper bound for $I_0(x) \exp(-x)$, e.g., $$ I_0(x) \exp(-x) \leq \frac{4}{3\sqrt{2\pi x+1}}. $$ Then $$ \frac{4}{3}(2n+1)\int_0^\infty(2\pi x+1)^{-(n+1)/2}=\frac{4}{3\pi}\frac{2n+1}{n-1} $$ and provided $n\geq 10$, $$ \psi(n+1)-\psi(n) \leq 0 $$


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