# Integral Evaluation with Partial Fractions

• J_M_R
In summary, the conversation discusses how to evaluate the integral I(alpha) = ∫1/((x+alpha^2)(x+1)) dx between the limits of 0 and infinity using partial fractions. It goes through the steps of finding the partial fractions and combining the natural logs to simplify the expression. It also addresses a special case for alpha = 1 and suggests using the limit to evaluate the integral. The conversation ends with a discussion on using L'Hopital's Rule and taking the limit to find the derivative of I(alpha).
J_M_R

## 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?

J_M_R said:

## 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?

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.

scurty said:
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)?

J_M_R said:
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.

J_M_R said:
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.

Don't forget to treat the special case of alpha=1.

J_M_R said:
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'.

## What is integral evaluation with partial fractions?

Integral evaluation with partial fractions is a technique used in calculus to simplify and evaluate integrals that involve rational functions. It involves breaking down a rational function into simpler fractions that can be more easily integrated.

## When is integral evaluation with partial fractions used?

This technique is typically used when the integrand (the function being integrated) is a rational function, meaning it is a fraction with polynomial expressions in the numerator and denominator.

## What are the steps for evaluating integrals using partial fractions?

The general steps for evaluating integrals with partial fractions are as follows:

1. Factor the denominator of the rational function into linear and irreducible quadratic factors.
2. Write the partial fraction decomposition, with each factor in the denominator represented by a separate fraction with an unknown constant in the numerator.
3. Find the values of the unknown constants by equating coefficients of like terms on both sides of the equation.
4. Use the decomposition to rewrite the integral as a sum of simpler integrals.
5. Integrate each simpler integral and combine the results to get the final answer.

## What are some common mistakes to avoid when using partial fractions for integration?

Some common mistakes to avoid when using partial fractions for integration include:

• Forgetting to account for all the factors in the denominator when writing the partial fraction decomposition.
• Incorrectly setting up equations to solve for the unknown constants.
• Forgetting to include the constant of integration when integrating each simpler integral.
• Making arithmetic errors when combining the results of the simpler integrals.

## Are there any other applications of partial fractions besides integration?

Yes, partial fractions are also commonly used in differential equations and in finding the inverse Laplace transform of a function. They can also be used in other areas of mathematics, such as in number theory and algebraic geometry.

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