Hot Integral - 11

Calculus Level 5

Let I n ( x ) = e 3 π i n / 2 2 n π ∫ − ∞ ∞ e − ( t − i x ) 2 t n d t \qquad \qquad \displaystyle I_n(x)=\frac{e^{3\pi in/2}2^n}{\sqrt{\pi}}\int\limits_{-\infty}^{\infty}e^{-(t-ix)^2}t^n dt

Then, ∑ n = 0 ∞ ( 2 a + 1 2 ) n I 2 n + 1 ( x ) y 2 n Γ ( 2 ( n + 1 ) ) = A ( y B + C ) − D a − E Q x ( X F Y ( a + G H ; J K ; y L x M y N + P ) ) \displaystyle \sum_{n=0}^{\infty} \frac{( \frac{2a+1}{2} )_n I_{2n+1}(x) y^{2n}}{\Gamma(2(n+1))} = A(y^B+C)^{- Da - \frac{E}{Q}} x (\;_XF_Y(a+\frac{G}{H} ; \frac{J}{K} ; \frac{y^Lx^M}{y^N+P}))

Calculate A + B + C + D + E + G + H + J + K + L + M + N + P + Q + X + Y A+B+C+D+E+G+H+J+K+L+M+N+P+Q+X+Y

Details and Assumptions:

- X F Y ( . . . ; . ; . . ) \;_XF_Y(...;.;..) represents hypergeometric functions.

- ( b ) n (b)_n represents Pochhammer symbol.

- gcd ⁡ ( E , Q ) = gcd ⁡ ( G , H ) = gcd ⁡ ( J , K ) = 1 \gcd(E,Q)=\gcd(G,H)=\gcd(J,K)=1

- A , B , C , D , E , G , H , J , K , L , M , N , P , Q , X , Y A,B,C,D,E,G,H,J,K,L,M,N,P,Q,X,Y are all positive integers.

- i = − 1 i=\sqrt{-1}

■ \blacksquare This is a part of Hot Integrals


The answer is 26.

Aman Rajput by Aman Rajput , 25, India

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1 solution

Mark Hennings
Feb 9, 2016

This is a combination of two fairly standard results. To begin with, this integral tells us that I n ( x ) = H n ( x ) I_n(x) = H_n(x) is the n n th Hermite polynomial. But then ∑ n = 0 ∞ ( 2 a + 1 2 ) n I 2 n + 1 ( x ) y 2 n Γ ( 2 n + 2 ) = ∑ n = 0 ∞ ( a + 1 2 ) n H 2 n + 1 ( x ) y 2 n ( 2 n + 1 ) ! = 2 ( y 2 + 1 ) − a − 1 2 x 1 F 1 ( a + 1 2 ; 3 2 ; x 2 y 2 y 2 + 1 ) \begin{array}{rcl} \displaystyle \sum_{n=0}^\infty \frac{\big(\frac{2a+1}{2}\big)_n I_{2n+1}(x) y^{2n}}{\Gamma(2n+2)} & = & \displaystyle \sum_{n=0}^\infty \frac{\big(a +\frac12\big)_n H_{2n+1}(x) y^{2n}}{(2n+1)!} \\ & = & \displaystyle 2(y^2+1)^{-a-\frac12} x \, {}_1F_1\big(a+\tfrac12\,;\,\tfrac32\,;\, \tfrac{x^2y^2}{y^2+1}\big) \end{array} using a less common generating function identity for the Hermite polynomials. Thus the answer is 2 + 2 + 1 + 1 + 1 + 1 + 2 + 3 + 2 + 2 + 2 + 2 + 1 + 2 + 1 + 1 = 26 . 2 + 2 + 1 + 1 + 1 + 1 + 2 + 3 + 2 + 2 + 2 + 2 + 1 + 2 + 1 + 1 \; = \; \boxed{26} \;.

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