Integration-Gaussian times polynomial

Calculus Level 4

If ∫ 0 ∞ e − x 2 ( 2 x + 1 ) 6 d x = A + B π 2 \displaystyle\int^{\infty}_0e^{-x^2}(2x+1)^6\ dx = A + \dfrac{B\sqrt{\pi}}{2} , find A + B A+B .


The answer is 609.

Max Wong by Max Wong , 18, Hong Kong

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1 solution

Mark Hennings
Nov 23, 2018

If I n = ∫ 0 ∞ x n e − x 2 d x I_n \; = \; \int_0^\infty x^n e^{-x^2}\,dx then I 0 = 1 2 π I_0 = \tfrac12\sqrt{\pi} , I 1 = 1 2 I_1 = \tfrac12 , while I n = 1 2 ( n − 1 ) I n − 2 n ≥ 2 I_n \; = \; \tfrac12(n-1)I_{n-2} \hspace{2cm} n \ge 2 so that I 2 = 1 4 π I_2 = \tfrac14\sqrt{\pi} , I 3 = 1 2 I_3 = \tfrac12 , I 4 = 3 8 π I_4 = \tfrac38\sqrt{\pi} , I 5 = 1 I_5 = 1 and I 6 = 15 16 π I_6 = \tfrac{15}{16}\sqrt{\pi} . Thus ∫ 0 ∞ e − x 2 ( 2 x + 1 ) 6 d x = 64 I 6 + 192 I 5 + 240 I 4 + 160 I 3 + 60 I 2 + 12 I 1 + I 0 = 278 + 331 2 π \int_0^\infty e^{-x^2}(2x+1)^6\,dx \; =\; 64I_6 + 192I_5 + 240I_4 + 160I_3 + 60I_2 + 12I_1 + I_0 \; = \; 278 + \tfrac{331}{2}\sqrt{\pi} making the answer 278 + 331 = 609 278 + 331 = \boxed{609} .

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