# Help for this improper integral

1. Mar 14, 2014

### 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 dont know how to find the antiderivative of the first fraction.

2. Mar 14, 2014

### jbunniii

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?

3. Mar 14, 2014

### Parhs

A hint was to use Laplace for this . I know your solution makes sense but I dont know how it should be written or if there is another way more formal

4. Mar 14, 2014

### jbunniii

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.

5. Mar 14, 2014

### Parhs

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

6. Mar 14, 2014

### scurty

Do you know how to perform trigonometric substitution?

7. Mar 14, 2014

### jbunniii

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.

8. Mar 15, 2014