For which value of $t$ does following integral converge $ \int_{0}^{\infty} \frac{x+1}{{3x^2-t}} - \frac{t}{{2x+1}} dx $?

Question

I am trying to find for which value of t, the following integral converges (and where it converges):

$$ \text {Let} \ t \in \mathbb{R}$$

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

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My attempt

Let's consider the second part of the integral only and by that I mean $$ \int_{0}^{\infty} \frac{t}{{2x+1}} dx $$ Now

$$\text{Let}\ u = 2x + 1$$

$$\text{Then}\ dx = \frac{1}{2} du $$ $$\text{and}\ \int_{0}^{\infty} \to \int_{1}^{\infty} $$

$$ \int_{1}^{\infty} \frac{t}{2u} du= \frac{t}{2}\left[ \ln(u)\right]^{+\infty}_1 = +\infty$$

Therefore it seems that this diverges, no matter what's the t value.
Therefore the initial integral (1) diverges too.
What is the mistake in this approach?

Answer 1

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

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

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

for this integral to converge, considering that it is a rational function we must have $deg(2x^2+3x+1 - 3tx^2+t^2) \leq deg((3x^2-t)(2x+1)) -2 = 1.$

Therefore $deg(2x^2+3x+1 - 3tx^2+t^2) \leq 1$ i.e : $2=3t$ or $t=\frac {2}{3}.$

Now we see what happens around our value of $t$:

$$\frac{2x^2+3x+1 - 3tx^2+t^2}{(3x^2-t)(2x+1)}$$ is equivalent to $$\frac{\sqrt2+1+\frac{4}{9}}{(3x^2-\frac{2}{3})(2(\sqrt2 /3)+1)}$$

which clearly is not convergent around $x=\frac{\sqrt2}{3}.$

In conclusion the integral does not converge ever

Answer 2

HINT :

You can write the integrand as $$ \frac{t^2-3tx^2+2x^2+3x+1}{\left(2x+1\right)\left(3x^2-t\right)} $$ So that you can prove there's no problem in 0 or in $+\infty$ by finding an equivalent which form is like $1/u^{\alpha}$.