From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Help me to take Integral. Date: 20 Jan 2000 07:04:46 GMT Newsgroups: sci.math Summary: Differentiating under integral signs to recognize a definite integral In article <01bf61bb$2cd81300$62b24c3e@pc1>, Alex Senkevich wrote: >Suggest, please, how to take an integral: > > infinity > / 2 > | x exp(-p x ) > | ------------ dx > | x + a > / > 0 > >[ x*exp(-p*x*x)/(x+a), x-from 0 to INFINITY ] > >p>0, a>0 (real) By using a linear change of variable, we may assume p = 1. It's a little easier to consider the function H(a) = (1/a) * integral[ x*exp(-x*x)/(x+a), x-from 0 to INFINITY ] One the one hand, using the fact that integral[ exp(-x*x), x=...]=sqrt(pi)/2, we get the reduction H(a) = sqrt(pi)/2/a - integral[ exp(-x*x)/(x+a), x-from 0 to INFINITY ] which we blissfully differentiate to see that H'(a) =-sqrt(pi)/2/a^2 + integral[ exp(-x*x)/(x+a)^2, x-from 0 to INFINITY ] On the other hand, we recognize the numerator in the integrand of H as a derivative, and so we integrate by parts to see 2a H(a) = 1/a - integral[ exp(-x*x)/(x+a)^2, x-from 0 to INFINITY ] Add to deduce H'(a) + 2 a H(a) = 1/a - sqrt(pi)/2/a^2. Thus the function K(a) = H(a) * exp(a*a) satisfies the equation K'(a) = exp(a*a) * ( 1/a - sqrt(pi)/2/a^2 ) You may thus express K in terms of "known" functions; maple renders this as / 2 \ 2 | exp(a ) 1/2 | 1/2 - 1/2 Ei(1, -a ) - 1/2 |- ------- - I Pi erf(I a)| Pi + C \ a / = -1/2*Ei(1,-a^2)-1/2*(-exp(a^2)/a-I*Pi^(1/2)*erf(I*a))*Pi^(1/2) + C but I don't know that this is really any easier than simply saying "K is the antiderivative of ...". Note that this "closed form" for K depends on knowing the value of K at some single point, in order to evaluate C. For example, we know K(1) = exp(1)*H(1), where H(1) -- which is your initial integral when a=1, p=1 -- is about .28109327294941343190. Plouffe's inverter did not recognize this number. Of course, given K(a), your original integral is then a*exp(-a*a)*K(a). dave