Does this integral converge at $y=1$?

[karush]

$\displaystyle
\int_{-\infty}^{0} \frac{1}{1+{x}^{2}}\,dx$

Due to the interval does this converge at $y=1$ ?

 

[topsquark]

$\displaystyle
\int_{-\infty}^{0} \frac{1}{1+{x}^{2}}\,dx$

Due to the interval does this converge at $y=1$ ?

You'll have to define y first. What is it?

-Dan

 

[karush]

$a=-\infty $ $b=0$
$\displaystyle
\int_{a}^{b} \frac{1}{1+{x}^{2}}\,dx$

$x=\tan\left({\theta}\right)
\therefore dx=\sec^2{\theta} \, d\theta
\therefore \theta = \arctan\left\{x\right\}$

$\displaystyle
\int_{a}^{b}\frac{\sec^2{\theta}}{\tan^2\left({\theta}\right)+1}
=\int_{a}^{b}\frac{\sec^2{\theta}}{\sec^2{\theta}} \,d\theta
=\int_{a}^{b}1\,d\theta
=\theta$
backsubstute $a=-\infty $ $b=0$ $\theta = \arctan\left\{x\right\}$
$=\left[\arctan\left(x\right) \right]_{-\infty}^0
=\frac{\pi}{2}$

 

[Greg]

That's correct, but what is $y$ ?

 

[karush]

$x=0,y=1$

But is that convergence
For $-\infty<x\le0$ ?

 

[topsquark]

$x=0,y=1$

But is that convergence
For $-\infty<x\le0$ ?

Yes, the integral converges on \(\displaystyle (-\infty, 0] \).

But what is y supposed to be? (Crying)

-Dan

 

[karush]

so you mean what $\frac{1}{1+{x}^{2}}$ approaches, not what it is at a point

as it goes to $-\infty$ it approaches $y=0$

 

[topsquark]

so you mean what $\frac{1}{1+{x}^{2}}$ approaches, not what it is at a point

as it goes to $-\infty$ it approaches $y=0$

So you are talking about the asymptote of the function 1/(1 + x^2). As x goes to - infinity then the asymptote is y = 0.

-Dan

 

[karush]

Don't know, hope so

 

[MarkFL]

If you are defining:

\(\displaystyle y(x)\equiv\frac{1}{x^2+1}\)

Then yes, we have:

\(\displaystyle y(0)=1\)

and:

\(\displaystyle \lim_{x\to\infty}y=0\)

However, the fact that the above limit is zero, is a necessary but not sufficient condition for the integral to converge. :D

 

[karush]

The domain is $x\le 0$ would that make it converge or is that the wrong direction

 

[MarkFL]

I did write the limit incorrectly, but we could note that an implication of the even function rule is that:

\(\displaystyle \int_{-\infty}^{0}\frac{1}{x^2+1}\,dx=\int_{0}^{\infty}\frac{1}{x^2+1}\,dx\)

What makes it converge isn't the domain but rather how quickly the integrand approaches zero as $x$ increases.