# A Definite integral where solution. involves infinity - infinity

1. Mar 18, 2009

### cpashok

Last edited: Mar 18, 2009
2. Mar 18, 2009

### Staff: Mentor

After using partial fractions decomposition, I get this:
$$\int_0^B \frac{-1/2}{x + 1} + (1/2)\frac{x + 1}{x^2 + 1}dx$$

I'm not sure this is what you got. After integrating, the antiderivative will be in terms of B. Take the limit of this expression as B approaches infinity.

3. Mar 18, 2009

### Dick

Combine the parts that go to infinity into one term. ln((x+1)^(1/2)/(x^2+1)^(1/4)). You want to figure out the limit of that as x->infinity. Can you see where to go from there?

4. Mar 18, 2009

### cpashok

Thank you very much Mark44 and Dick.
I got the same integral as Mark44.
I understand what Dick is saying but I'm not allowed to use L hospital's rule at school. I can't figure out any other way.
Is there an alternate way to evaluate the above integral?

5. Mar 18, 2009

### Dick

Factor an x^(1/2) out of the numerator and denominator of (x+1)^(1/2)/(x^2+1)^(1/4).

6. Mar 18, 2009

### cpashok

Correct me if i am wrong.
ln [x^(1/2) { 1+ 1/ x^(1/2) } / x^(1/2) { 1+ 1/ x^(1/2) }]

I would get ln 1.

which is equal to zero and hence the required answer.
Thanks a lot Mark 44 and Dick.

7. Mar 18, 2009

### Dick

The conclusion is right, but your algebra leaves a lot to be desired. If you mean
(x+1)^(1/2)=(x)^(1/2)*(1+1/x)^(1/2), that's ok. If you mean (x+1)^(1/2)=x^(1/2)+1/(x^(1/2)) or x^(1/2)+(1/x)^(1/2), both of those are wrong. And what happened with the 1/4 power in the denominator? Try to use enough parentheses to make it clear what you mean.

8. Mar 18, 2009

### cpashok

oops!
{x^(1/2){1+1/x}^(1/2)} / x ^(1/2){1+1/(x^2)}^(1/4) .
i got this in the notebook, forgot to use paranthesis and forgot to edit the second term that i copy pasted from the first.

9. Mar 18, 2009

Perfect.