Hello!
I am trying to evaluate the following integral:
$\displaystyle \int_{-\infty}^{0}(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}})\,dx$
The result is supposed to be ln2.
Could somebody please help me with this?
Hey kareman.
Try using this:
Hyperbolic function - Wikipedia, the free encyclopedia
Since
$\displaystyle \frac{1}{\sqrt{x^2+a^2}}<\frac{1}{\sqrt{x^2}}$ the integral $\displaystyle \int_{-\infty}^{0}\frac{1}{\sqrt{x^2+a^2}}\,dx$ diverges because the integral $\displaystyle \int_{-\infty}^{0}\frac{1}{\sqrt{x^2}}\,dx$ diverges.
The same applies to $\displaystyle \int_{-\infty}^{0}\frac{1}{\sqrt{x^2+4a^2}}$.
Therefore it is not possible to use the formula $\displaystyle \int\frac{1}{\sqrt{x^2+a^2}}\,dx=\ln(x+\sqrt{x^2+a ^2})$ for each fraction separately for this improper integral.
Perhaps, I am missing something else here?
Yes... if I proceed this way then for x=0:
$\displaystyle \ln(x+\sqrt{x^2+a^2})=\ln(a)$ for the first fraction and $\displaystyle \ln(x+\sqrt{x^2+4a^2})=\ln(2a)$ for the second fraction which equals to -ln2, NOT ln2.
However the result should be ln2, so what am I doing wrong?
Given arcsin(u) = ln(u + SQRT(u^2 + 1)) we get a cancellation for x = 0 but for x = -infinity we should get a non-zero term when everything is cancelled.
We solve the limit of lim x->-infinity [(x/a) + SQRT((x/a)^2 + ]/[(x/4a) + SQRT((x/4a)^2 + 1))] and then take the log of that number.
OK, after struggling too much time with this, I tried again with a clear mind. Here is my calculation (I have checked the LaTeX code, but the symbols for the square root (in two cases) and for the fraction (in one case) don't show up normally):
$\displaystyle \int_{-\infty}^{0}(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}})\,dx=$
$\displaystyle \lim_{b\rightarrow-\infty}[\ln{(x+\sqrt{x^2+a^2}})-\ln{(x+\sqrt{x^2+4a^2)}}]|_b^0=$
$\displaystyle \lim_{b\rightarrow-\infty}[\ln{|a|}-\ln{|2a|}-ln{(b+\sqrt{b^2+a^2}})+\ln{(b+\sqrt{b^2+4a^2)}}]=$
$\displaystyle -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{(b+\sqrt{b^2+4a^2)}}-ln{(b+\sqrt{b^2+a^2}})]=$
$\displaystyle -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{(b+\sqrt{b^2+4a^2})}{(b+\sqrt{b^2+a^2})} }]=$
$\displaystyle -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{b+\sqrt{b^2(1+\frac{4a^2}{b^2})}}{b+\sqr t{b^2(1+\frac{a^2}{b^2})}}]=$
$\displaystyle -\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{b+|b|\sqrt{(1+\frac{4a^2}{b^2})}}{b+|b|\ sqrt{(1+\frac{a^2}{b^2})}}]$
b goes to $\displaystyle -\infty$ therefore $\displaystyle |b|=-b$, so
$\displaystyle -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{b-b\sqrt{(1+\frac{4a^2}{b^2})}}{b-b\sqrt{(1+\frac{a^2}{b^2})}})}=$
$\displaystyle -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{1-\sqrt{(1+\frac{4a^2}{b^2})}}{1-\sqrt{(1+\frac{a^2}{b^2})}})}$
the limit is in the indeterminate form $\displaystyle \frac{0}{0}$ so by using L'Hospital's rule I get
$\displaystyle -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{-\frac{1}{2\sqrt{(1+\frac{4a^2}{b^2})}}(1+\frac{4a^ 2}{b^2})'}{-\frac{1}{2\sqrt{(1+\frac{a^2}{b^2})}}(1+\frac{a^2} {b^2})'}=$
$\displaystyle -\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{\frac{1}{2\sqrt{(1+\frac{4a^2}{b^2})} }(4a^2(-2b^{-1}))}{\frac{1}{2\sqrt{(1+\frac{a^2}{b^2})}}(a^2(-2b^{-1}))}=$
$\displaystyle -\ln{2}+\ln{\lim_{b\rightarrow-\infty}4\frac{\sqrt{1+\frac{a^2}{b^2}}}{\sqrt{1+\f rac{4a^2}{b^2}}}=$
$\displaystyle -\ln{2}+\ln{4}=$
$\displaystyle -\ln{2}+\ln{2^2}=$
$\displaystyle -\ln{2}+2\ln{2}=$
$\displaystyle \ln{2}$
Alternatively, it occured to me that the function $\displaystyle \frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}}$ is an even function, therefore the integral from $\displaystyle -\infty$ to 0 has the same value as the integral from 0 to $\displaystyle \infty$, which is easier to calculate (without using L'Hospital's rule).
@MINOAMAN: Yes, I am Greek.