Improper integral with parameter calculation

Question

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 ?

score 2 · Answer 1 · answered Mar 13 '14 at 13:31

Look at the asymptotics. For $x \to \infty$,

$$\frac{x+1}{3x^2+\lambda} \sim \frac{1}{3x},\tag{1}$$

regardless of $\lambda$, and

$$\frac{\lambda}{2x+1} \sim \frac{\lambda}{2x}.\tag{2}$$

Thus

$$\frac{x+1}{3x^2+\lambda} - \frac{\lambda}{2x+1} = \frac{1}{3x} - \frac{\lambda}{2x} + O\left(\frac{1}{x^2}\right)$$

for $x\to \infty$. Hence the integral can only be finite if - what?

score 1 · Accepted Answer · answered Mar 13 '14 at 13:41

1

$$\frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1} = \frac{x^2(2-3\lambda)+3x + 1 - \lambda^2}{6x^3 +3x^2 + 2\lambda x + \lambda}$$

If $\lambda \neq \frac{2}{3}$, then the above ratio is asymptotic to $\frac{3x}{6x^3} = \frac{1}{2x^2}$ and hence the integral has a finite solution.

If $\lambda = \frac{2}{3}$, then the above ratio is asymptotic to $\frac{x^2(2-3\lambda)}{6x^3} = \frac{2-3\lambda}{6x}$ and hence the integral has a infinite solution.

answered Mar 13 '14 at 13:41

the_candyman

14,064
4
35
62

How did you find the asymptotic? – GorillaApe Mar 13 '14 at 13:53
When you have something in the form $$\frac{\sum_{i=0}^n a_ix^i}{\sum_{j=0 }^m b_j x^j}$$, with $a_n \neq 0$ and $b_m \neq 0$, then it is asymptotic to $$\frac{a_n x^n}{b_m x^m}$$. – the_candyman Mar 13 '14 at 15:56

score 1 · Answer 3 · edited Mar 13 '14 at 14:43

Just as a modest complement to Daniel Fisher's answer, the antiderivative (for positive $\lambda$) is $$\frac{1}{6} \log \left(3 x^2+\lambda\right)-\frac{1}{2} \lambda \log (2 x+1)+\frac{\tan ^{-1}\left(\frac{\sqrt{3} x}{\sqrt{\lambda }}\right)}{\sqrt{3} \sqrt{\lambda }}$$ The integral for $x=0$ to $a$ is $$\frac{1}{6} \left(\log \left(3 a^2+\lambda \right)-3 \lambda \log (2 a+1)+\frac{2 \sqrt{3} \tan ^{-1}\left(\frac{\sqrt{3} a}{\sqrt{\lambda }}\right)}{\sqrt{\lambda }}-\log (\lambda )\right)$$ from which you can conclude in a more complex manner to the same as from Daniel Fisher's answer.

Just for your legitimate curiosity, the Taylor expansion for large values of $a$ is given by $$\frac{1}{6} \left(3 \lambda \log \left(\frac{1}{a}\right)-2 \log \left(\frac{1}{a}\right)+\sqrt{3} \pi \sqrt{\frac{1}{\lambda }}-3 \lambda \log (2)-\log (\lambda )+\log (3)\right)+\frac{-\frac{\lambda }{4}-\frac{1}{3}}{a}+O\left(\left(\frac{1}{a}\right)^2\right)$$

Improper integral with parameter calculation

3 Answers3