1. ## Improper Integral

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.

3. ## Re: Improper Integral

chiro thank you for your reply. I tried this way but I don't get the result ln2...

4. ## Re: Improper Integral

What result do you get?

5. ## Re: Improper Integral

I can't evaluate each integral that involves a fraction separately because I get limits that do not exist. Each integral that involves a fraction diverges.

6. ## Re: Improper Integral

That should not happen if a is non-zero.

Show us what you have tried.

7. ## Re: Improper Integral

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?

8. ## Re: Improper Integral

You've bought up a great point, but do remember that ln(x) - ln(y) = ln(x/y) (Try using that to get some cancellations).

9. ## Re: Improper Integral

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?

10. ## Re: Improper Integral

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.

11. ## Re: Improper Integral

but I don't know how to type math symbols to show you the result if anyone wants..please let me know....
Minoas
P.S Careman are you a Greek?

12. ## Re: Improper Integral

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.