I want to calculate the integration of this function:
$$ \int_{-\infty }^\infty { 1\over \sqrt{(x-a)^2 + (y-b)^2}} dy $$ where $ a $ and $ b $ are the constants. The integral comes out to be $$ \ln \left ( \left | \sqrt{(x-a)^2 + (y-b)^2} + y -b \right | \right ) + C $$ Which is divergent in nature over the limits. I have following questions:
Is there any other way to deal with such integrals?
- Is there any way to find any approximate function of this which does not have this divergence issue?
I am quite new to this stuff, so sorry for any mistakes in this post. thanks for any help in advance.
The above question is already answered by Dr. MV, i have a query while following the same rule in second half of my equation : $$ \int_{-\infty }^\infty { y\over \sqrt{(x-a)^2 + (y-b)^2}} dy $$ the integral of this comes out to be $$ \ln \left ( \left | \sqrt{(x-a)^2 + (y-b)^2} + y -b \right | \right ) + \sqrt {(x-a)^2 + (y-b)^2} + C $$
this function still remains divergent. is there any way to deal with this
thanks for any help
The integral of interest fails to exist even as a Cauchy Principal Value. To see this, we write
$$\begin{align} \text{PV}\left(\int_{-\infty }^\infty { 1\over \sqrt{(x-a)^2 - (y-b)^2}} dy\right)&=\lim_{L\to \infty}\int_{-L}^L \frac{1}{\sqrt{(x-a)^2+(y-b)^2}}\,dy\\\\ &=\lim_{L\to \infty} \left. \left(\log\left((y-b)+\sqrt{(x-a)^2+(y-b)^2}\right)\right)\right|_{-L}^L\\\\ &=\lim_{L\to \infty}\log\left(\frac{(L-b)+\sqrt{(x-a)^2+(L-b)^2}}{-(L+b)+\sqrt{(x-a)^2+(L+b)^2}}\right)\\\\ &= \infty \end{align}$$
If we wish to analyze the behavior of the integral before passing to the limit, we can proceed as follows.
$$\begin{align} \log\left(\frac{(L-b)+\sqrt{(x-a)^2+(L-b)^2}}{-(L+b)+\sqrt{(x-a)^2+(L+b)^2}}\right)&=\log\left(\frac{(L-b)\left(\sqrt{1+\left(\frac{x-a}{L-b}\right)^2}+1\right)}{(L+b)\left(\sqrt{1+\left(\frac{x-a}{L+b}\right)^2}-1\right)}\right)\\\\ &\sim 2\log(L) \end{align}$$