Two integrals of Ramanujan
Two integrals of Ramanujan
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DOI：
10.1090/s0002-9939-1982-0652440-2
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发表时间：
1982-02
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影响因子：
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通讯作者：
R. Askey
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文献类型：
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作者：
R. Askey
计算了Ramanujan的两个积分。在1976年G. Andrews发现的Ramanujan恒等式页[2J]中，有一页是关于正规积分的积分。下面给出了拉马努金恒等式的推导。为了便于打印，设置q e-2k2和00 (1)(a;q)00 = 1（1aq '）。Ramanujan给出了以下恒等式：I e-z2+2mz(-ae2kzq; q)oo(be-2kZq; q)00 dx (2) a) Vr(abq; q)ooem e-2k2 (aql/2e2mk; q)oo(bql/2e-2mk; q)oo q = 0？-z -2+2mx dx (3) Jo (aql/2e2ikx; q)00(bq1/ 2e2ikx; q)00 = m (- aq1 /2e2ikx; q)00(- bq1 -2imk;q) 00 _ 21c (abq; q)00 q=e为了得到这些结果，我们将使用q-二项式定理00(a;q)n n (ax; q)00 q|<1, |xI<1，以及它(1)nq(n2_n)/2xn (5) n=O (qq) =(x; q)00, I qj< 1，其中(6)(a;q)n =(a;q) O/(aqn; q)00。参见[1，定理2.11，[31]或[4]，第661页的证明。(2)和(3)中的参数m可以通过a和b的平移和重定义去除，因此我们假设m = 0。为了证明(2)在每个有限产品上的使用(5)编辑于1981年7月31日收到。1980数学学科分类。主要33 a15。
TWO integals of Ramanujan are evaluated. In the pages of identities of Ramanujan that G. Andrews found in 1976 [2J, there is one page of integrals related to the normal integral. Derivations of Ramanujan's identities are given below. For ease in printing, set q e-2k2 and 00 (1) (a;q)00 = 1(1 aq'). n=O Ramanujan stated the following identities: I e-z2+2mz(-ae2kzq; q)oo(be-2kZq; q)00 dx (2) a ) Vr(abq; q)ooem e-2k2 (aql/2e2mk; q)oo(bql/2e-2mk; q)oo q = 0? -ze-2+2mx dx (3) Jo (aql/2e2ikx; q)00(bq1/2e-2ikx; q)00 = m (-aqe 2mk; q)OO(-bqe-2imk ; q)00 _ 21c (abq; q)00 q=e To obtain these results we will use the q-binomial theorem 00 (a;q)n n (ax; q)00 q|<1, |xI<1, and a limiting case of it (1)nq(n2_n)/2xn (5) n=O (qq) =(x; q)00, I qj< 1 where (6) (a; q)n = (a; q)OO/(aqn; q)00. See [l, Theorem 2.11, [31 or [4, p. 661 for proofs. The parameter m in (2) and (3) can be removed by translation and redefinition of a and b, so we assume m = 0. To prove (2) use (5) on each of the ifinite products Received by the editors July 31, 1981. 1980 Mathematics Subject Cassification. Primary 33A15.