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MATHEMATICS
O. S. BERLYAND and A. Ya. PRESSMAN
ASYMPTOTIC REPRESENTATIONS AND SOME ESTIMATES FOR INTEGRAL ERROR FUNCTIONS OF ARBITRARY ORDER
(Presented by Academician A. A. Dorodnitsyn, 21 III 1961)
The solution of many problems of mathematical physics is conveniently expressed through \(i^\mu \operatorname{erfc} z\)—integral error functions of order \(\mu\), which are solutions of the equation
\[ w''+2zw'-2\mu w=0 \tag{1} \]
and, for \(\operatorname{Re}\mu>-1\), admit the representation in the form of the integral
\[ i^\mu \operatorname{erfc} z= \frac{2}{\sqrt{\pi}}\frac{e^{-z^2}}{\Gamma(1+\mu)} \int_0^\infty t^\mu e^{-2zt-t^2}\,dt . \tag{2} \]
Here we shall consider some properties of the function \(i^\mu \operatorname{erfc} z\), or, for brevity, \(i_\mu(z)\), for real values of the index and argument.
- \(\mu=\nu>-1;\ z=x\ge 0\). This case has been treated in detail in \((^{1-3})\), where expressions have been obtained both for small and for large \(x\):
\[ i_\nu(x)= \frac{e^{-x^2}}{\sqrt{\pi}\,\Gamma(1+\nu)} \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}(2x)^k \Gamma\!\left(\frac{1+\nu+k}{2}\right); \tag{3} \]
\[ i_\nu(x)= \frac{2e^{-x^2}}{\sqrt{\pi}\,(2x)^{\nu+1}} \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \frac{\Gamma(1+\nu+2k)}{\Gamma(1+\nu)} (2x)^{-2k}. \tag{4} \]
If both the argument and the index increase simultaneously, while \(\nu/x\sqrt{2}\ll 1\), then, using the method set forth in (4), we obtain
\[ i_\nu(x)\sim \frac{2e^{-x^2-\nu^2/4x^2}}{\sqrt{\pi}\,(2x)^{\nu+1}} \left[ 1+\frac{1}{6\nu} \left( 1-11\frac{\nu^2}{2x^2} +3\frac{\nu^4}{4x^4} \right) \right]. \tag{5} \]
From the recurrence relations
\[ 2\nu i_\nu(x)=-2x i_{\nu-1}(x)+i_{\nu-2}(x); \tag{6} \]
\[ i'_\nu(x)=-i_{\nu-1}(x) \tag{7} \]
we estimate the modulus of the logarithmic derivative \(i_{\nu-1}(x)/i_\nu(x)\). From (6) we have
\[ \frac{i_{\nu-1}(x)}{i_\nu(x)} = 2x+\frac{2(\nu+1)}{2x+i_\nu(x)/i_{\nu+1}(x)} \leq 2x+\frac{2(\nu+1)}{2x}. \]
On the other hand,
\[ \frac{i_\nu(x)}{i_{\nu+1}(x)} = 2x+ \frac{2(\nu+2)} {2x+\dfrac{2(\nu+3)}{2x+i_{\nu+2}(x)/i_{\nu+3}(x)}} \geq 2x+ \frac{2(\nu+2)} {2x+\dfrac{2(\nu+3)}{2x}}, \]
whence, for \((2x)^2\geq(\nu+1)(\nu+3)\), we obtain
\[ i_\nu^2(x)\geq i_{\nu-1}(x)i_{\nu+1}(x) \quad \text{for } \nu\geq -1. \tag{8} \]
- \(\mu=\nu>-1;\ z=-x\leqslant 0\). In this case
\[ i_{\nu}(-x)=\frac{2}{\sqrt{\pi}\Gamma(1+\nu)}e^{-x^{2}}\int_{0}^{\infty}t^{\nu}e^{2xt-t^{2}}\,dt =\frac{2}{\sqrt{\pi}\Gamma(1+\nu)}x^{\nu}\int_{-x}^{\infty}\left[1+\frac{\tau}{x}\right]^{\nu}e^{-\tau^{2}}\,d\tau . \tag{9} \]
Writing \(e^{2xt}=1+2xt+(2xt)^2/2!+\cdots\), we obtain
\[ i_{\nu}(-x)=\frac{1}{\sqrt{\pi}\Gamma(1+\nu)}e^{-x^{2}} \sum_{k=0}^{\infty}\frac{(2x)^k}{k!}\, \Gamma\!\left(\frac{1+\nu+k}{2}\right). \tag{10} \]
The asymptotic expansion of \(i_{\nu}(-x)\) for large \(x\) will be found from (9), if we use the facts that
\[ \int_{-x}^{+x}\tau^{2s+1}e^{-\tau^{2}}\,d\tau=0 \quad\text{and}\quad \int_{-x}^{+x}\tau^{2s}e^{-\tau^{2}}\,d\tau =\frac{\sqrt{\pi}(2s)!}{s!\,4^s}+O(e^{-x^{2}}). \]
Then
\[ i_{+\nu}(-x)=\frac{2}{\Gamma(1+\nu)}x^{\nu} \left\{ \sum_{k=0}^{N-1} \frac{\Gamma(1+\nu)}{k!\,\Gamma(1+\nu-2k)}(2x)^{-2k} +O\!\left[(2x)^{-2N}\right]+R_{\nu} \right\}, \tag{11} \]
where \(R_{\nu}<\sqrt{\pi/2}\,(2/x)^{\nu}\operatorname{erfc}(x/\sqrt{2})\).
For integral \(\nu\) the series in (11) terminates and becomes, for \(\nu=2k\),
\[ i_{2k}(-x)=\frac{2}{(2k)!}x^{2k} \left\{ \sum_{s=0}^{k}\frac{(2k)!}{s!(2k-2s)!}(2x)^{-2s} +R_{2k} \right\} \tag{11a} \]
and for \(\nu=2k+1\),
\[ i_{2k+1}(-x)=\frac{2}{(2k+1)!}x^{2k+1} \left\{ \sum_{s=0}^{k}\frac{(2k+1)!}{s!(2k+1-2s)!}(2x)^{-2s} +R_{2k+1} \right\}. \tag{11b} \]
Further, from (6) we have
\[ \frac{i_{\nu-1}(-x)}{i_{\nu}(-x)} = \frac{2\nu}{2x+i_{\nu-2}(-x)/i_{\nu-1}(-x)} \leqslant \frac{2\nu}{2x}, \]
since for \(\nu\geqslant 1\), \(i_{\nu-2}(-x)/i_{\nu-1}(-x)>0\). On the other hand,
\[ \frac{i_{\nu}(-x)}{i_{\nu+1}(-x)} = \frac{2(\nu+1)} {2x+\dfrac{2\nu}{\,2x+i_{\nu-2}(-x)/i_{\nu-1}(-x)\,}} \geqslant \frac{2(\nu+1)}{2x+\dfrac{2\nu}{2x}}, \]
whence, for \((2x)^2\geqslant 2\nu^2\), we obtain
\[ i_{\nu}^{2}(-x)\geqslant i_{\nu-1}(-x)i_{\nu+1}(-x). \tag{12} \]
We note that (12) is also valid for \(\nu\geqslant -1\), which is established directly from (9) and (11).
We turn to functions with negative index less than \(-1\). Using the analytic continuation of \(\Gamma(1+\mu)\) into the region \(\operatorname{Re}\mu\leqslant -1\)*, we shall everywhere in the coefficients of the expansions of paragraphs 1 and 2 replace \(+\nu\) by \(-\nu\). Then we obtain results for the following case:
- \(\mu=-\nu\leqslant -1;\ z=x\geqslant 0\). From expressions (3) and (4) we obtain
\[ i_{-\nu}(x)=\frac{1}{\sqrt{\pi}\Gamma(1-\nu)}e^{-x^{2}} \sum_{k=0}^{\infty}\frac{(-1)^k}{k!}(2x)^k \Gamma\!\left(\frac{1-\nu+k}{2}\right); \tag{13} \]
\[ i_{-\nu}(x)=\frac{2e^{-x^{2}}}{\sqrt{\pi}}(2x)^{\nu-1} \sum_{k=0}^{\infty}\frac{(-1)^k}{k!}(2x)^{-2k} \frac{\Gamma(1-\nu+2k)}{\Gamma(1-\nu)}. \tag{14} \]
\[ \text{* For all complex }\mu\text{ }(^{5}) \]
\[ \Gamma(1+\mu)= \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\frac{1}{1+\mu+n} +\int_{1}^{\infty}t^{\mu}e^{-t}\,dt. \]
For integer \(\nu\), the series in (13) and (14) terminate because of the poles of the \(\Gamma\)-functions, and we obtain the well-known relation between \(i_{-n}(x)\) and the Hermite polynomial
\[ i_{-n}(x)=\frac{2}{\sqrt{\pi}}e^{-x^{2}}H_{n-1}(x). \tag{15} \]
Using the asymptotic expansion (14) for \(2x>\nu+1\), one may obtain
\[ i_{-\nu}^{2}(x)\geq i_{-\nu-1}(x)i_{-\nu+1}(x) \quad \text{for }-\nu\leq -1. \tag{16} \]
- \(\mu=-\nu\leq -1;\ z=-x\leq 0\). In this case the expressions (10) and (11) of item 2, after changing the sign of \(\nu\), become
\[ i_{-\nu}(-x)=\frac{1}{\sqrt{\pi}\Gamma(1-\nu)}e^{-x^{2}} \sum_{k=0}^{\infty}\frac{(2x)^k}{k!}\, \Gamma\left(\frac{1-\nu+k}{2}\right); \tag{17} \]
\[ i_{-\nu}(-x)= \frac{2x^{-\nu}}{\Gamma(1-\nu)} \left\{ \sum_{k=0}^{N-1} \frac{\Gamma(1-\nu)}{k!\Gamma(1-\nu-2k)}(2x)^{-2k} +O\left[(2x)^{-2N}\right]+R_{-\nu} \right\}, \tag{18} \]
where \(R_{-\nu}>\sqrt{\pi}\,2^{-(1+\nu)}\operatorname{erfc}x\).
For integer \(\nu\), (17) also reduces to (15), but now for a negative argument. The asymptotic behavior of \(i_{-n}(-x)\) for integer \(n\) can be found from (14), replacing there \(x\) by \(-x\): for large \(x\), \(i_{-n}(-x)\) tends to zero as
\[ i_{-n}(-x)\sim \frac{2}{\sqrt{\pi}}e^{-x^{2}}(-2x)^{\,n-1}. \tag{19} \]
As for fractional \(\nu\), it follows from (18) that \(i_{-\nu}(-x)\) asymptotically tends to zero as
\[ i_{-\nu}(x)\sim \frac{2}{\Gamma(1-\nu)}x^{-\nu}. \tag{20} \]
We note that, because of the pole of \(\Gamma(1-n)\) for integer \(n\), for \(i_{-n}(-x)\) we immediately obtain 0 from (18) and (20). From the same relations we find that, asymptotically, \(i_{-\nu-1}(-x)/i_{-\nu}(-x)\sim -\nu/x\) for \((2x)^2>(\nu+1)(\nu+2)\), whence
\[ i_{-\nu}^{2}(-x)\geq i_{-\nu-1}(-x)i_{-\nu+1}(-x), \quad \nu \text{ noninteger }\geq 1. \tag{21} \]
If \(\nu=n\) (\(n\) integer), then from (16) and (18), for \(2x>n+1\), we find
\(i_{-n-1}(-x)/i_{-n}(-x)=-i_{-n-1}(x)/i_{-n}(x)\sim -2x\), whence
\[ i_{-n}^{2}(-x)\leq i_{-n-1}(-x)i_{-n+1}(-x). \tag{22} \]
Further, under the corresponding restrictions on \(x\) from (6), (8), and (12), we obtain
\[ \frac{\nu}{\nu+1}\leq \frac{i_{\nu-1}(x)i_{\nu+1}(x)}{i_{\nu}^{2}(x)} \leq 1 \quad \text{for } \nu\geq -1. \tag{23} \]
In conclusion we note that relations analogous to (16), (21), (22) can also be obtained for the functions \(H_n(x)\) and \(D_\nu(x)\), if one uses (15) and the expression
\[ i_\nu(x)=\frac{2}{\sqrt{\pi}}e^{-x^{2}/2}\cdot 2^{-\frac{\nu+1}{2}}D_{-\nu-1}(x\sqrt{2}). \]
Received
20 III 1961
CITED LITERATURE
- E. T. Whittaker, G. N. Watson, A Course of Modern Analysis, 2, 1934.
- W. Magnus, F. Oberhettinger, Formeln und Sätze für die speziellen Funktionen, Leipzig, 1948.
- O. S. Berlyand, L. A. Sokolovskaya, Engineering-Physics Journal, 1, No. 11 (1958).
- A. Ya. Pressman, O. S. Berlyand, DAN, 126, No. 3, 508 (1959).
- V. I. Smirnov, A Course of Higher Mathematics, 3, 1950.