Integration of an infinite product

ecpietscheck
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Hey guys, what sup
I need you all to help me in resolving the integral of an infinite product...
i was thinking of perhaps integrating by parts, but when yo do that the integration becomes brutally expansive...
any ideas?
thank you all very much

the variable which is aimed to be integrated is x btw...
 

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You wrote down a finite product. Do you in fact want ##\int \prod_{m=1}^\infty (x^2 + m) \, dx##?
 
for example

(x*x+1)*(x*x+2)*(x*x+3)*(x*x+4) = x^8 + a*x^6 + b*x^4 + c*x^2 + 1*2*3*4

1*2*3*4 = 4!


(x*x+1)*(x*x+2)*(x*x+3)*(x*x+4) = x^8 + 10x^6 + 35x^4 + 50x^2 + 24

Each member must be integrated.

∫ x^8 + 10x^6 + 35x^4 + 50x^2 + 24 dx =

= x^9/9 + 10x^7/7 + 7x^5 +50x^3/3 +24x + C

kamke
 
kamke said:
(x*x+1)*(x*x+2)*(x*x+3)*(x*x+4) = x^8 + 10x^6 + 35x^4 + 50x^2 + 24

Each member must be integrated.
The problem is to get a general formula for those coefficients.

Basically ##\displaystyle \sum_{i,j,...=1, i<j<...}^n i*j*k*...## with 0 to n indices in the sum.
The second sum, for example, sums 1*2+1*3*...+1*n + 2*3+2*4+...+2*n+...
 
MFB, that answer seems a bit unclear
perhaps any other ideas?
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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