# Definite Integral

• May 14th 2008, 06:25 PM
bloomsgal8
Definite Integral
I am having trouble with this definite integral, especially since it involves infinity, and the way my professor explained the process has me completely baffled. Any help is very much appreciated, last time one of the members of this forum explained the process better than my professor! Thank you for any help !

P.s. I haven't figured out how to display the definite integral with the math tag sorry about that...

Definite Integral (a= 2 and b=oo infinity) of $1/(x^2-1) dx$
• May 14th 2008, 06:38 PM
PaulRS
Note that: $\frac{1}{x^2-1}=\frac{1}{2}\cdot\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$

Thus: $\int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\cdot \lim_{b\rightarrow{+\infty}}\left(\int_2^b\frac{dx }{x-1}-\int_2^b\frac{dx}{x+1}\right)$

$\int_2^b\frac{dx}{x-1}=\ln(b-1)$ and $\int_2^b\frac{dx}{x+1}=\ln(b+1)-\ln(3)$

Thus: $\lim_{b\rightarrow{+\infty}}\left(\int_2^b\frac{dx }{x-1}-\int_2^b\frac{dx}{x+1}\right)=\ln(3)+\lim_{b\right arrow{+\infty}}\ln\left(\frac{b-1}{b+1}\right)$

But: $\lim_{b\rightarrow{+\infty}}\ln\left(\frac{b-1}{b+1}\right)=0$ since $\frac{b-1}{b+1}=\frac{1-\frac{1}{b}}{1+\frac{1}{b}}\rightarrow{1}$ and the logarithm is a continuous function.

Thus: $\int_2^{\infty}\frac{dx}{x^2-1}=\frac{\ln(3)}{2}$
• May 14th 2008, 06:42 PM
Mathstud28
Quote:

Originally Posted by bloomsgal8
I am having trouble with this definite integral, especially since it involves infinity, and the way my professor explained the process has me completely baffled. Any help is very much appreciated, last time one of the members of this forum explained the process better than my professor! Thank you for any help !

P.s. I haven't figured out how to display the definite integral with the math tag sorry about that...

Definite Integral (a= 2 and b=oo infinity) of $1/(x^2-1) dx$

$\int_2^{\infty}\frac{dx}{x^2-1}$

Using either the definition of $arctanh(x)$

Or doing PFD

to get

$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$

$1=A(x+1)+B(x-1)$

Letting $x=1$ we see that $A=\frac{-1}{2}$

and letting $x=-1$ we see that $B=\frac{-1}{2}$

So $\frac{1}{x^2-1}=\frac{-1}{2}\bigg[\frac{1}{x-1}-\frac{1}{x+1}\bigg]$

So $\int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\int_2^{\infty}\bigg[\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{2}\bigg[\ln(x+1)-\ln(x-1)\bigg]\bigg|_2^{\infty}=\frac{\ln(3)}{2}$
• May 14th 2008, 06:45 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
$\int_2^{\infty}\frac{dx}{x^2-1}$

Using either the definition of $arctanh(x)$

Or doing PFD

to get

$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$

$1=A(x+1)+B(x-1)$

Letting $x=1$ we see that $A=\frac{1}{2}$

and letting $x=-1$ we see that $B=\frac{-1}{2}$

So $\frac{1}{x^2-1}=\frac{1}{2}\bigg[\frac{1}{x-1}-\frac{1}{x+1}\bigg]$

So $\int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\int_2^{\infty}\bigg[\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{2}\bigg[\ln(x+1)-\ln(x-1)\bigg]\bigg|_2^{\infty}=\infty=\frac{\ln(3)}{2}$

And as I said alternatively $\int_2^{\infty}\frac{dx}{x^2-1}=arctanh(x)\bigg
|_2^{\infty}=\frac{\ln(3)}{2}$
• May 14th 2008, 06:58 PM
topsquark
Quote:

Originally Posted by Mathstud28
$\int_2^{\infty}\frac{dx}{x^2-1}=\frac{1}{2}\int_2^{\infty}\bigg[\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{2}\bigg[\ln(x+1)-\ln(x-1)\bigg]\bigg|_2^{\infty}=\infty=\frac{\ln(3)}{2}$

Confuscious say: There something wrong in last line. (Doh)

-Dan
• May 14th 2008, 07:14 PM
hatsoff
Quote:

Originally Posted by bloomsgal8
I am having trouble with this definite integral, especially since it involves infinity, and the way my professor explained the process has me completely baffled. Any help is very much appreciated, last time one of the members of this forum explained the process better than my professor! Thank you for any help !

P.s. I haven't figured out how to display the definite integral with the math tag sorry about that...

Definite Integral (a= 2 and b=oo infinity) of $1/(x^2-1) dx$

Well, as you might expect, we should first begin by finding the antiderivative of the function. So:

$F(x)=\int\frac{dx}{x^2-1}$

$\frac{1}{x^2-1}=\frac{1}{2}(\frac{1}{x-1}-\frac{1}{x+1})$

$F(x)=\frac{1}{2}\int\frac{dx}{x-1}-\frac{1}{2}\int\frac{dx}{x+1}$

$F(x)=\frac{1}{2}ln \frac{x-1}{x+1}$

Now, recall the formula for definite integrals:

$\int_{a}^{b}f(x)dx=F(b)-F(a)$

Therefore:

$\int_{2}^{\infty}f(x)dx=F(\infty)-F(2)$

$F(\infty)-F(2)=[\frac{1}{2}ln \frac{\infty-1}{\infty+1}]-[\frac{1}{2}ln \frac{2-1}{2+1}]$

$F(\infty)-F(2)=ln \sqrt{3}$