Posted by March 4, 2018
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! > > Gene
It does thanks. Sorry about the typo in the question, should have been a ln there as previous poster pointed out
Posted by Evgeny Filatov March 4, 2018
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
Posted by Marcel Mueller March 4, 2018
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
Posted by March 3, 2018
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.