# Can not get Integral Right

1. Feb 10, 2014

### muzialis

Hi All,

I met the following function to evaluate,
$$v(x)=\int_{-\infty}^{-a} G(t) ln(\vert t - x \vert) \mathrm{d}t + \int_{-a}^{a} -N ln(\vert t - x \vert) \mathrm{d}t + \int_{a}^{\infty} G(t) ln(\vert t - x \vert) \mathrm{d}t$$, where G is an unknown even function, N is a constant.
After multiple attempts I discovered I am not even close to the solution, which is achieved by the change of variables $$t = ar \qquad x = a \zeta \qquad g(r) = G(ar)/N$$, leading to the solution
$$aN \{2ln (a) \int_{1}^{\infty} g(r) \mathrm{d}r + \int_{1}^{\infty} g(r) ln (\vert r^{2} - \zeta^{2} \vert ) -2 ln (a) -ln (\vert \zeta^{2}-1 \vert) + \zeta \, ln (\frac{\zeta - 1}{\zeta + 1}) +2 \}$$,

wellI get only the first term, but am completely lost as to where the squared variables in the logarithm argument came from in the second term, where could they come from...

Any hint would be the most welcome.
thanks as usual

2. Feb 10, 2014

### pasmith

Since $g$ is even, you have
$$\int_{-\infty}^{-1} g(r) \ln|r - \zeta| \,dr = \int_1^\infty g(r) \ln |r + \zeta|\,dr$$
and adding $\int_1^\infty g(r) \ln |r - \zeta|\,dr$ and simplifying the logarithms will give the second term.

3. Feb 10, 2014

### muzialis

Thank you ever so much. That is massive progress. The third term, $$-2ln(a)$$ is easily explained, all there is to do now is to understand the origin of the fourth and fifth terms...
They should originate from the integral
$$\int_{-a}^{a} ln (\vert r - \zeta \vert) \mathrm{d}r$$
which I should be able to compute by splitting it into $$\int_{-a} ^{\zeta} \quad \int_{\zeta}^{a}$$in order to get rid of the absolute value and then use a basic antiderivative, I am on it now..
thanks again