MHB Is the Integral \int^0_{-\infty} \frac{1}{3 - 4x} dx Divergent or Convergent?

[shamieh]

Determine whether the integral is Divergent or Convergent$$\int^0_{-\infty} \frac{1}{3 - 4x} dx$$

I did a u substitution and got

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

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

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

 

[Prove It]

Determine whether the integral is Divergent or Convergent$$\int^0_{-\infty} \frac{1}{3 - 4x} dx$$

I did a u substitution and got

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

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

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

Yes it is divergent.

 

[MarkFL]

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

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

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

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

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

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

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

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

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

Thus, the given integral is divergent.