# Convergence and Divergence

#### shamieh

##### Active member
Determine whether the integral is Divergent or Convergent

$$\displaystyle \int^0_{-\infty} \frac{1}{3 - 4x} dx$$

I did a u substitution and got

$$\displaystyle \lim_{a\to\infty} -\frac{1}{4}\sqrt{3} + \frac{1}{4}\sqrt{3 - 4a}$$

So is because the $$\displaystyle -\infty$$ is under the square root is it going to be divergent?

I have $$\displaystyle \frac{1}{4}\sqrt{3 - 4\infty}$$

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#### Prove It

##### Well-known member
MHB Math Helper
Determine whether the integral is Divergent or Convergent

$$\displaystyle \int^0_{-\infty} \frac{1}{3 - 4x} dx$$

I did a u substitution and got

$$\displaystyle \lim_{a\to\infty} -\frac{1}{4}\sqrt{3} + \frac{1}{4}\sqrt{3 - 4a}$$

So is because the $$\displaystyle -\infty$$ is under the square root is it going to be divergent?

I have $$\displaystyle \frac{1}{4}\sqrt{3 - 4\infty}$$
Yes it is divergent.

#### MarkFL

Staff member
It is divergent, but not for the reason you cite. I would first use the substitution:

$$\displaystyle u=3-4x\,\therefore\,du=-4\,dx$$ and our integral becomes:

$$\displaystyle I=-\frac{1}{4}\int_{\infty}^3\frac{du}{u}$$

Applying the rule $$\displaystyle -\int_b^a f(x)\,dx=\int_a^b f(x)\,dx$$ we obtain:

$$\displaystyle I=\frac{1}{4}\int_3^{\infty}\frac{du}{u}$$

Now, since this is an improper integral, we may write:

$$\displaystyle I=\frac{1}{4}\lim_{t\to\infty}\left[\int_3^t\frac{du}{u} \right]$$

Applying the FTOC, and a property of logarithms, we find:

$$\displaystyle I=\frac{1}{4}\lim_{t\to\infty}\left(\ln\left(\frac{t}{3} \right) \right)=\infty$$

Thus, the given integral is divergent.