# Bound for a Confluent hypergeometric function ${}_1F_1$

When I have $$f(a,N,x)={}_1F_1\left(1;N;a\frac{x}{1+x}\right)$$ where $$N$$ positive integer, $$a>0$$ and $$x\in[0,\infty)$$, I may have a bound $$f(a,N,x)\leq{}_1F_1\left(1;N;a\right)$$ as $$\frac{x}{1+x}\in[0,1]$$. This is good enough for my work.

However, now, I have $$f(a,N,x)={}_1F_1\left(1;N;a x\right)$$ which cannot be bounded as previous case.

Does anyone have idea of getting some bound such as $$f(a,N,x)\leq g(a,N) h(a,N,x)$$ where parameter $$x$$ in $$h(a,N,x)$$ may be only in polynomial or exponential functions, e.g., $$x^pe^{-qx}$$, $$x^pe^{-qx^2}$$.

I'm unsure of what is 'good enough' for your work, but one such bound arises as follows:

First, we rewrite this particular special case of Kummer's (confluent hypergeometric) function in terms of the lower incomplete gamma function:

$$_1F_1(1;N;ax) = (N-1)(ax)^{1-N}e^{ax}\gamma(N-1,ax).$$

This is a well-known identity, e.g., lower incomplete gamma

Next, we can use the following bound

$$(N-1)(ax)^{1-N}\gamma(N-1,ax) \leq \frac{1}{N}(1 + (N-1)e^{-ax}).$$

(this comes from the paper Inequalities and Bounds for the Incomplete Gamma Function)

Combining these results in the bound:

$$_1F_1(1;N;ax) \leq \frac{e^{ax}}{N}(1 + (N-1)e^{-ax}) = \frac{e^{ax}-1}{N} + 1.$$

Hopefully this will suffice for your purposes.

• Is it possible to use $e^a$ instead $e^{ax}$? Thanks !
– Frey
May 11, 2017 at 9:13
• Because i really need $e^{-px}$ form or else $x^p$ only which is the most preferred one.
– Frey
May 11, 2017 at 9:23
• @frey What ranged of values do you have for your inputs? May 11, 2017 at 11:20