I have an integral 1/(a^2-x^2) wrt x. If you use partial fractions we get (1/2a) * ln [ (x+a)/(x-a) ] +c and zero initial conditions gives c=0 however, this is a standard integral and should be (1/a)arctanh(x/a) How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x) when I use this the sign is wrong.

# Simple Integration

On 04.03.18 02.13, gyansorova@gmail.com wrote:> I have an integral > > 1/(a^2-x^2) wrt x. If you use partial fractions we get > > (1/2a) * ln [ (x+a)/(x-a) ] +c>> however, this is a standard integral and should be > > (1/a)arctanh(x/a) > > How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x)The latter is wrong. arctanh(x) := ln((1+x)/(1-x))/2 And this will make the above terms equivalent. Marcel

On 04.03.2018 4:13, gyansorova@gmail.com wrote:> I have an integral > > 1/(a^2-x^2) wrt x. If you use partial fractions we get > > (1/2a) * ln [ (x+a)/(x-a) ] +c > > and zero initial conditions gives c=0 > > however, this is a standard integral and should be > > > (1/a)arctanh(x/a) > > How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x) > > when I use this the sign is wrong. >Gyansorova, the integral 1/(a^2-x^2) wrt x is only defined for regions (1) x < -a, (2) -a < x < a, and (3) x > a. You cannot include points x = -a and x = a within the integration limits, because that would be improper integral. Without involving complex calculus, the best option is to use Cauchy principal value as a common workaround for dealing with improper integrals. However, as long as we are going to stay with real and proper integrals, you are limited to either of the regions -- (1), (2) or (3). Now, remember that the indefinite integral of 1/x is not ln(x) + C, as many people seem to mistakenly assume, but ln(|x|) + C, where |x| denotes the absolute value of x. Which means that if you use partial fractions, for regions (1) and (3) you get (1/2a) * ln [ (x+a)/(x-a) ] + C. While for region (2) you get (1/2a) * ln [ - (x+a)/(x-a) ] + C. In the same time, arctanh(x) as a real function is defined for -1<x<1 -- because tanh(x) belongs to the region (-1; 1) for any real x. Which means that the expression (1/a) arctanh(x/a) + C is only defined for region (2), where it coincides with the answer you get using partial fractions. Hope that helps! Gene

On Sunday, March 4, 2018 at 11:54:54 PM UTC+13, Evgeny Filatov wrote:> On 04.03.2018 4:13, gyansorova@gmail.com wrote: > > I have an integral > > > > 1/(a^2-x^2) wrt x. If you use partial fractions we get > > > > (1/2a) * ln [ (x+a)/(x-a) ] +c > > > > and zero initial conditions gives c=0 > > > > however, this is a standard integral and should be > > > > > > (1/a)arctanh(x/a) > > > > How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x) > > > > when I use this the sign is wrong. > > > > Gyansorova, the integral 1/(a^2-x^2) wrt x is only defined for regions > (1) x < -a, (2) -a < x < a, and (3) x > a. > > You cannot include points x = -a and x = a within the integration > limits, because that would be improper integral. Without involving > complex calculus, the best option is to use Cauchy principal value as a > common workaround for dealing with improper integrals. > > However, as long as we are going to stay with real and proper integrals, > you are limited to either of the regions -- (1), (2) or (3). > > Now, remember that the indefinite integral of 1/x is not ln(x) + C, as > many people seem to mistakenly assume, but ln(|x|) + C, where |x| > denotes the absolute value of x. > > Which means that if you use partial fractions, for regions (1) and (3) > you get (1/2a) * ln [ (x+a)/(x-a) ] + C. While for region (2) you get > (1/2a) * ln [ - (x+a)/(x-a) ] + C. > > In the same time, arctanh(x) as a real function is defined for -1<x<1 -- > because tanh(x) belongs to the region (-1; 1) for any real x. Which > means that the expression (1/a) arctanh(x/a) + C is only defined for > region (2), where it coincides with the answer you get using partial > fractions. > > Hope that helps! > > GeneIt does thanks. Sorry about the typo in the question, should have been a ln there as previous poster pointed out