, If a and b are finite, we can reduce to

, If a (or b) is finite and the function has an algebraic singularity at a (or b), we remove the singularity by a polynomial change of variable

, 0 (say) and b = ?, then if f does not tend to 0 exponentially fast (for instance f (x) ? 1/x k )

, If a = 0 (say) and b = ? and if f does tend to 0 exponentially fast (for instance f (x) ? e ?ax or f (x) ? e ?ax 2 ), we use x = ? (t) = exp

?. , use x = ? (t) = sinh(sinh(t)) if f does not tend to 0 exponentially fast

, the quotient-difference algorithm compute c(m) such that ?(z) = c(0)z/(1 + c(1)z/(1 + c(2)z/(1 + · · ·))) (see the remark made above in case the algorithm has a division by 0; it may also happen that the odd or even moments vanish, so that the continued fraction is only in powers of z 2, Set ?(z) = ? k?0 M k z k+1 , and using

, For any m, denote as usual by p m (z)/q m (z) the mth convergent obtained by stopping the continued fraction at c(m)z/1, and denote by N n (z) the reciprocal polynomial of p 2n?1 (z)/z (which has degree n ? 1) and by D n (z) the reciprocal polynomial of q 2n?1

, The x i are the n roots of D n (which are all simple and in the interval ]a, b[), and the w i are given by the formula w i = N n

, By construction, this Gaussian integration method will work when the function f (x) to be integrated is well approximated by polynomials, but otherwise will fail miserably, and this is why we say that the method is much less "robust" than doublyexponential integration. The fact that Gaussian

, /n) k /n 2 = ? (k + 2), so that ?(z) = ? k?1 ? (k + 1)z k. Note that this is closely related to the digamma function ?(z), but we do not need this. Applying the quotient-difference algorithm, we write ?(z) n?1 f (1/n)/n 2 , or equivalently (changing the definition of f ) that ? i w i f (y i ) is a very good approximation to ? n?1 f (n), f (x)dµ = ? n?1 f (1/n)/n 2. Let us apply the recipe given above: the kth moment M k is given by M k = ? n?1

·. , ·. , and W. , = 10.3627543 · · ·, and (by definition) we have ? 1?i?2 w i f (y i ) =

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