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Exercise: In a similar manner, compute 1 x 0 (e − 1 − x)/x2 dx. From now on, we assume by splitting the interval that we have only one singularity, at the beginning of the interval, which by translation we may assume to be [0, 1]. If the singularity is removable or of logarithmic type, we proceed as above. On the other hand, if the singularity is algebraic, we must tell the program: for example, working with the default accuracy of 38 decimal digits, intnum(x=0,1,1/sqrt(x)) has 30 correct digits, but intnum(x=0,1,x^(-3/4)) only has 15.

15. vv,vv=intnumainfinit(1)); [b,h,vabs,vwt,vabsneg,vwtneg]=vv; N=#vabs; e1=exp(-1); b=b; if (a<1, return ((2*e1*f(a+e1)+sum(m=1,N,vwt[m]\ *f(a+vabs[m])+vwtneg[m]*f(a+vabsneg[m])))*h), return ((2*e1*f(a*(1+e1))+sum(m=1,N,vwt[m]\ *f(a*(1+vabs[m]))+vwtneg[m]*f(a*(1+vabsneg[m]))))*a*h); ); } (Note: the b=b command above is to avoid a compiler warning if compiled with gp2c). 1. (1) In the built-in function intnum, an integral such ∞ ∞ as a f (t) dt is simply treated as 0 f (a + t) dt, which is mathematically correct, but induces errors when a is very large.

Let us look in more detail at these two methods. For the first, recall that with our assumptions on F we have a decomposition into partial fractions of the form aα,k F (x) = , (x − α)k α pole 1≤k≤−v(α) where −v(α) ≥ 1 denotes the order of the pole α of f , and α αa,1 = 0 since the degree of F is less than or equal to −2. From this it is immediate to see that ∞ aα,k F (x) dx = − aα,1 log(N − α) + . , around 0, so if N is large enough we have ∞ N F (x) dx = m≥2 bm /((m − 1)N m−1 ) . To be able to use this formula, we need an estimate on the growth of bm so as to see where we can truncate the series.

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A most easy method for finding many very large prime numbers by Euler L.


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