Improper Integral

**kareman** · February 6th 2013, 02:30 PM

Hello!

I am trying to evaluate the following integral:

$\int_{-\infty}^{0}(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}})\,dx$

The result is supposed to be ln2.

Could somebody please help me with this?

**chiro** · February 6th 2013, 02:38 PM

Hey kareman.

Try using this:

Hyperbolic function - Wikipedia, the free encyclopedia

**kareman** · February 6th 2013, 03:42 PM

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

**chiro** · February 6th 2013, 03:43 PM

What result do you get?

**kareman** · February 6th 2013, 03:49 PM

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.

**chiro** · February 6th 2013, 05:47 PM

That should not happen if a is non-zero.

Show us what you have tried.

**kareman** · February 7th 2013, 12:16 AM

Since
$\frac{1}{\sqrt{x^2+a^2}}<\frac{1}{\sqrt{x^2}}$ the integral $\int_{-\infty}^{0}\frac{1}{\sqrt{x^2+a^2}}\,dx$ diverges because the integral $\int_{-\infty}^{0}\frac{1}{\sqrt{x^2}}\,dx$ diverges.

The same applies to $\int_{-\infty}^{0}\frac{1}{\sqrt{x^2+4a^2}}$ .

Therefore it is not possible to use the formula $\int\frac{1}{\sqrt{x^2+a^2}}\,dx=\ln(x+\sqrt{x^2+a ^2})$ for each fraction separately for this improper integral.

Perhaps, I am missing something else here?

**chiro** · February 7th 2013, 12:27 AM

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).

**kareman** · February 7th 2013, 02:05 AM

Yes... if I proceed this way then for x=0:

$\ln(x+\sqrt{x^2+a^2})=\ln(a)$ for the first fraction and $\ln(x+\sqrt{x^2+4a^2})=\ln(2a)$ for the second fraction which equals to -ln2, NOT ln2.

However the result should be ln2, so what am I doing wrong?

**chiro** · February 7th 2013, 02:23 PM

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.

**MINOANMAN** · February 7th 2013, 10:23 PM

The answer is ln2 .
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?

**kareman** · February 8th 2013, 09:54 AM

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):

$\int_{-\infty}^{0}(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}})\,dx=$

$\lim_{b\rightarrow-\infty}[\ln{(x+\sqrt{x^2+a^2}})-\ln{(x+\sqrt{x^2+4a^2)}}]|_b^0=$

$\lim_{b\rightarrow-\infty}[\ln{|a|}-\ln{|2a|}-ln{(b+\sqrt{b^2+a^2}})+\ln{(b+\sqrt{b^2+4a^2)}}]=$

$-\ln{2}+\lim_{b\rightarrow-\infty}[\ln{(b+\sqrt{b^2+4a^2)}}-ln{(b+\sqrt{b^2+a^2}})]=$

$-\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{(b+\sqrt{b^2+4a^2})}{(b+\sqrt{b^2+a^2})} }]=$

$-\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{b+\sqrt{b^2(1+\frac{4a^2}{b^2})}}{b+\sqr t{b^2(1+\frac{a^2}{b^2})}}]=$

$-\ln{2}+\lim_{b\rightarrow-\infty}[\ln{\frac{b+|b|\sqrt{(1+\frac{4a^2}{b^2})}}{b+|b|\ sqrt{(1+\frac{a^2}{b^2})}}]$

b goes to $-\infty$ therefore $|b|=-b$ , so

$-\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{b-b\sqrt{(1+\frac{4a^2}{b^2})}}{b-b\sqrt{(1+\frac{a^2}{b^2})}})}=$

$-\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{1-\sqrt{(1+\frac{4a^2}{b^2})}}{1-\sqrt{(1+\frac{a^2}{b^2})}})}$

the limit is in the indeterminate form $\frac{0}{0}$ so by using L'Hospital's rule I get

$-\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{-\frac{1}{2\sqrt{(1+\frac{4a^2}{b^2})}}(1+\frac{4a^ 2}{b^2})'}{-\frac{1}{2\sqrt{(1+\frac{a^2}{b^2})}}(1+\frac{a^2} {b^2})'}=$

$-\ln{2}+\ln{\lim_{b\rightarrow-\infty}\frac{\frac{1}{2\sqrt{(1+\frac{4a^2}{b^2})} }(4a^2(-2b^{-1}))}{\frac{1}{2\sqrt{(1+\frac{a^2}{b^2})}}(a^2(-2b^{-1}))}=$

$-\ln{2}+\ln{\lim_{b\rightarrow-\infty}4\frac{\sqrt{1+\frac{a^2}{b^2}}}{\sqrt{1+\f rac{4a^2}{b^2}}}=$

$-\ln{2}+\ln{4}=$

$-\ln{2}+\ln{2^2}=$

$-\ln{2}+2\ln{2}=$

$\ln{2}$

Alternatively, it occured to me that the function $\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{\sqrt{x^2+4a^2}}$ is an even function, therefore the integral from $-\infty$ to 0 has the same value as the integral from 0 to $\infty$ , which is easier to calculate (without using L'Hospital's rule).

@MINOAMAN: Yes, I am Greek.

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