# Help for this improper integral

• Parhs
In summary, the student is trying to find the value of λ for which the integrand would return a number (not infinity) and is having trouble because he doesn't know how to perform trigonometric substitution. After some experimentation, he finds that the solution is 2/3 and plugs it back into the original integrand to simplify it. He then proves that the result is integrable.
Parhs
Let $$λ \in R$$

$$I=\int_{0}^{\infty} \left(\frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1}\right)dx$$

I need to find λ for which this would return a number (not infinity) . I tried writing Numerators as derivatives but not sure about the correctness and results.eg $$\fracλ2\int\frac{d(2x+1)}{2x+1}$$

Any idea how to solve this ?

I don't know how to find the antiderivative of the first fraction.

When ##x## is large,
$$\frac{x+1}{3x^2 + \lambda} \approx \frac{1}{3x}$$
and
$$\frac{\lambda}{2x+1} \approx \frac{\lambda}{2x}$$
so
$$\frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1} \approx \frac{1/3 - \lambda/2}{x}$$
Can you use these approximations to determine what ##\lambda## must be, given the fact that the "tail" of ##1/x## is too "heavy" to be integrated?

A hint was to use Laplace for this . I know your solution makes sense but I don't know how it should be written or if there is another way more formal

Parhs said:
A hint was to use Laplace for this . I know your solution makes sense but I don't know how it should be written or if there is another way more formal
Do you mean Laplace transforms? I'm not sure how they would be useful here, but probably there's some equivalent formulation that uses them.

If you solve for the value of ##\lambda## using the approach I outlined, you can simply replace it back into the original integrand and simplify it. You should be able to easily verify that the result is integrable.

Hm but is this approach correct ? (how can I proof that this is correct ?And put this simpler version within integral ?)
Here is how the book does it
http://i.imgur.com/59IHuKQ.jpg

Parhs said:
I don't know how to find the antiderivative of the first fraction.

Do you know how to perform trigonometric substitution?

Well, I can't read the Greek, and I'm not sure how they are going to solve for ##\lambda## after the last line in the image. Even if it works, the method seems very complicated compared with the straightforward one I suggested.

As I said, to see that my method is correct, first solve for the value of ##\lambda## that will eliminate the problematic ##1/x##: obviously you want ##1/3 - \lambda / 2 = 0##, so ##\lambda = 2/3##.

Then to see that this value of ##\lambda## works, just plug it into the original integrand and simplify it:

$$\frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1} = \frac{x+1}{3x^2 + 2/3} - \frac{2/3}{2x+1}$$
I'll let you do the algebra, but the key is that the ##x## term in the numerator will vanish, leaving you with an integrand which is asymptotically a multiple of ##1/x^2## as ##x \rightarrow \infty## (and bounded as ##x \rightarrow 0##), hence integrable. No other value of ##\lambda## will do this for you.

## 1. What is an improper integral?

An improper integral is an integral where either the upper or lower limit of integration is infinite, or the integrand function is undefined at one or more points in the interval of integration.

## 2. How do I evaluate an improper integral?

To evaluate an improper integral, you must first determine if it is convergent or divergent. If it is convergent, you can use various methods such as the limit comparison test, the comparison test, or the limit comparison test to find its value. If it is divergent, the integral does not have a finite value.

## 3. What are some common types of improper integrals?

Some common types of improper integrals include integrals with infinite limits, integrals with discontinuous integrands, and integrals with singularities (points where the integrand is undefined).

## 4. How do I determine if an improper integral is convergent or divergent?

To determine if an improper integral is convergent or divergent, you can use the limit comparison test, the comparison test, or the limit comparison test. These tests involve comparing the improper integral to a known convergent or divergent integral.

## 5. What are some applications of improper integrals?

Improper integrals have many applications in physics, engineering, and economics. They are used to model real-world situations where the limits of integration may be infinite or the integrand may have discontinuities or singularities. They are also used in advanced calculus and analysis to evaluate complex functions and solve differential equations.

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