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Integral Evaluation with Partial Fractions

  • Thread starter J_M_R
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  • #1
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Homework Statement



I(alpha) = ∫1/((x+alpha^2)(x+1)) dx between the limits of 0 and infinity

Evaluate the integral above depending on the parameter alpha using partial fractions.


The Attempt at a Solution



1/((x+alpha^2)(x+1)) = A/(x+alpha^2) + B/(x+1)

1 = A(x+1) + B(x+alpha^2)

let x = -1 then B = 1/(-1+alpha^2)

let x = (-alpha^2) then A = 1/(-alpha^2+1)

therefore the integral becomes two integrals that can be added together:

1/(-alpha^2+1)∫1/(x+alpha^2)dx + 1/(-1+alpha^2)∫1/(x+1)dx

both integrals are definite integrals between 0 and infinity

When I integrate 1/(x+1) between zero and infinity I obtain infinity as an answer. This is the same as when I integrate the other partial fraction 1/(x+alpha^2) between zero and infinity. Therefore when I multiply this by the constants that I have already taken outside of the integral my answer to the evaluation of the integral will be infinity.

This seems to be correct as when the x value increases, the graph will not reach a zero point so as this continues forever, the area under the graph will be equal to infinity.

What action do I need to take from the point of having the partial fractions in order to evaluate this integral properly?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
392
17

Homework Statement



I(alpha) = ∫1/((x+alpha^2)(x+1)) dx between the limits of 0 and infinity

Evaluate the integral above depending on the parameter alpha using partial fractions.


The Attempt at a Solution



1/((x+alpha^2)(x+1)) = A/(x+alpha^2) + B/(x+1)

1 = A(x+1) + B(x+alpha^2)

let x = -1 then B = 1/(-1+alpha^2)

let x = (-alpha^2) then A = 1/(-alpha^2+1)

therefore the integral becomes two integrals that can be added together:

1/(-alpha^2+1)∫1/(x+alpha^2)dx + 1/(-1+alpha^2)∫1/(x+1)dx

both integrals are definite integrals between 0 and infinity

When I integrate 1/(x+1) between zero and infinity I obtain infinity as an answer. This is the same as when I integrate the other partial fraction 1/(x+alpha^2) between zero and infinity. Therefore when I multiply this by the constants that I have already taken outside of the integral my answer to the evaluation of the integral will be infinity.

This seems to be correct as when the x value increases, the graph will not reach a zero point so as this continues forever, the area under the graph will be equal to infinity.

What action do I need to take from the point of having the partial fractions in order to evaluate this integral properly?

Homework Statement





Homework Equations





The Attempt at a Solution

When you get to the part with all the natural logs, instead of taking the limit at this point, try combining all the natural logs into one and using l'Hospital's Rule.
 
  • #3
20
0
When you get to the part with all the natural logs, instead of taking the limit at this point, try combining all the natural logs into one and using l'Hospital's Rule.
Thank you,

I have combined the logs to get:

1/(1-alpha^2) (ln((x+alpha^2)/(x+1))) with the limits 0 and infinity around the ln section, the first part is the constant as before.

When the infinite limit is applied we get ln((infinity +alpha^2)/(infinity +1)) so I used L'hopital's rule:

If lim(x → c) f(x)/g(x) = L then, lim(x → c) f'(x)/g'(x) = L

For my question...

f(x) = x+alpha^2 f'(x) = alpha^2
g(x) = x+1 g'(x) = 1

so my integral becomes:

1/(1-alpha^2) [ln(alpha^2/1] with the x limits around []

= 1/(1-alpha^2) [lnalpha^2-lnalpha^2]
= 1/(1-alpha^2) [ln1]
= 0

Is this correct? I am sure I have made a mistake as I now need to use my result to find I'(alpha)?
 
  • #4
392
17
Thank you,

I have combined the logs to get:

1/(1-alpha^2) (ln((x+alpha^2)/(x+1))) with the limits 0 and infinity around the ln section, the first part is the constant as before.

When the infinite limit is applied we get ln((infinity +alpha^2)/(infinity +1)) so I used L'hopital's rule:

If lim(x → c) f(x)/g(x) = L then, lim(x → c) f'(x)/g'(x) = L

For my question...

f(x) = x+alpha^2 f'(x) = alpha^2
g(x) = x+1 g'(x) = 1

so my integral becomes:

1/(1-alpha^2) [ln(alpha^2/1] with the x limits around []

= 1/(1-alpha^2) [lnalpha^2-lnalpha^2]
= 1/(1-alpha^2) [ln1]
= 0

Is this correct? I am sure I have made a mistake as I now need to use my result to find I'(alpha)?
I won't be near a computer until later tonight, but after integrating you should have 4 terms. When you plug in 0 to one of the terms, it turns to zero and goes away. Now you are left with three terms. For one of the terms you need to factor out a -1 to make the coefficients all line up. Then you can coming the natural logs and then take the limit. Perhaps someone else can write this out more clearly, if not I will reply later tonight.
 
  • #5
392
17
1/(-alpha^2+1)∫1/(x+alpha^2)dx + 1/(-1+alpha^2)∫1/(x+1)dx

both integrals are definite integrals between 0 and infinity

When I integrate 1/(x+1) between zero and infinity I obtain infinity as an answer. This is the same as when I integrate the other partial fraction 1/(x+alpha^2) between zero and infinity. Therefore when I multiply this by the constants that I have already taken outside of the integral my answer to the evaluation of the integral will be infinity.
Okay, so you have ##\displaystyle\lim_{a\to\infty} \left[\frac{1}{1-\alpha^2}\int_0^a \frac{1}{x+\alpha^2} dx + \frac{1}{\alpha^2-1} \int_0^a \frac{1}{x+1} dx\right]## and then you correctly obtained natural logarithms from the integrations:

##\displaystyle\lim_{a\to\infty} \left[\frac{1}{1-\alpha^2} ln(x+\alpha^2) + \frac{1}{\alpha^2-1} ln(x+1)\right]_0^a##

Before you take the limit as a goes to infinity, plug in the a's and 0's and combine the natural logs into one natural log. You'll have a common factor of ##\frac{1}{1-\alpha^2}## and you can thus combine the logs into one. When you do, you can finally take the limit with no complications.
 
  • #6
18
2
Don't forget to treat the special case of alpha=1.
 
  • #7
vela
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I have combined the logs to get:

1/(1-alpha^2) (ln((x+alpha^2)/(x+1))) with the limits 0 and infinity around the ln section, the first part is the constant as before.

When the infinite limit is applied we get ln((infinity +alpha^2)/(infinity +1)) so I used L'hopital's rule:
I'll assume you did the integration correctly. You don't need to use the Hospital rule here. Instead, just evaluate the limit the normal way:
$$\lim_{x \to \infty} \frac{x+\alpha^2}{x+1} = \lim_{x \to \infty} \frac{1+\frac{\alpha^2}{x}}{1+\frac1x} = ?$$

If lim(x → c) f(x)/g(x) = L then, lim(x → c) f'(x)/g'(x) = L

For my question...

f(x) = x+alpha^2 f'(x) = alpha^2
g(x) = x+1 g'(x) = 1
If you insist on using the Hospital rule, make sure you differentiate correctly. Check f'.
 

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