A Definite integral where solution. involves infinity - infinity

In summary: Thank you for helping to solve the problem.In summary, Mark44 and Dick attempted to solve the integral equation x^(1/2) + 1/(x^(1/2)) = 0, but were unable to do so using standard algebraic methods. They were then able to find an alternate solution using partial fractions decomposition.
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  • #2
After using partial fractions decomposition, I get this:
[tex]
\int_0^B \frac{-1/2}{x + 1} + (1/2)\frac{x + 1}{x^2 + 1}dx
[/tex]

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
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
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
cpashok said:
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?

Factor an x^(1/2) out of the numerator and denominator of (x+1)^(1/2)/(x^2+1)^(1/4).
 
  • #6
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
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
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
cpashok said:
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.

Perfect.
 

What is a definite integral where the solution involves infinity - infinity?

A definite integral where the solution involves infinity - infinity is a type of integral where the upper and lower limits of integration result in an infinite value. This can happen when the function being integrated has a vertical asymptote at one or both of the limits, causing the integral to be undefined.

How do you solve a definite integral where the solution involves infinity - infinity?

The solution to a definite integral where the solution involves infinity - infinity can be found by using one of two methods: the first is to evaluate the integral as if it were a Cauchy principal value, which involves taking the limit as the upper and lower limits approach the point of discontinuity. The second method is to use a change of variables to transform the integral into one that is well-defined.

What are some common functions that result in a definite integral where the solution involves infinity - infinity?

Functions that have vertical asymptotes at the limits of integration, such as 1/x or tan(x), often result in a definite integral where the solution involves infinity - infinity. Other functions, such as e^x or ln(x), may also result in this type of integral if the limits are chosen in a way that causes the function to become undefined.

Why is a definite integral where the solution involves infinity - infinity important?

A definite integral where the solution involves infinity - infinity is important because it highlights the limitations of the definite integral as a tool for finding the area under a curve. It also allows for the study of functions that have vertical asymptotes, which are commonly encountered in real-world applications.

What are some real-world applications of a definite integral where the solution involves infinity - infinity?

A definite integral where the solution involves infinity - infinity can be used in various fields such as physics, engineering, and economics. For example, in physics, it can be used to calculate the work done by a force on an object that moves along a path with a vertical asymptote. In economics, it can be used to model supply and demand curves that have vertical asymptotes at certain prices.

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