- #1

nobahar

- 495

- 2

Hello!

I have a couple of questions on the following. Firstly, I was hoping someone could check my working and my reasoning.

Secondly, I was wondering if someone knew an alternative way of solving this problem.

I wanted to integrate this from [itex]x = -\infty[/itex] to [itex]x = \infty[/itex]:

[tex]\lim_{a \rightarrow 0^{\pm}} \left(\frac{a}{\pi(x^2 + a^2)}\right)[/tex]

It looks kind of like the standard Cauchy distribution-thingy...

I figured since [itex]a \neq 0[/itex] then I can ignore the limit (##a## could just be a really small number, for example, or a large number, and, as a constant, except for the sign it wouldn't make a difference what its value is). I put [itex]a \rightarrow 0^{\pm}[/itex] rather than just ##a \rightarrow 0## because the limit is different from either side, which is also important. I think...

Therefore:

[tex]\int_{-\infty}^{\infty} \left( \lim_{a \rightarrow 0^{\pm}} \left(\frac{a}{\pi(x^2 + a^2)}\right) \right) dx = \lim_{a \rightarrow 0^{\pm}} \left( \int_{-\infty}^{\infty}\frac{a}{\pi(x^2 + a^2)} dx \right)[/tex]

I used a trig substitution [itex]x = a \ tan(\theta)[/itex] since the range of ##\theta## is the same as x: [itex]-\infty < \tan(\theta) < \infty[/itex]. To change the endpoints to ##\theta##, for a>0 I get the lower bound to be [itex]\frac{-\pi}{2}[/itex] and the upper bound to be [itex]\frac{\pi}{2}[/itex], but for ##a<0## I get ##\frac{\pi}{2}## and ##\frac{-\pi}{2}## (they 'flip'); for the example, using ##a>0##:

[tex]\int_{-\infty}^{\infty}\frac{a}{(x^2 + a^2)} dx = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{a}{\pi((a \tan(\theta))^2 + a^2)}\,a (\tan^2(\theta) + 1) d\theta = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{a^2 (\tan^2(\theta) + 1)}{\pi a^2 (\tan^2(\theta) + 1)} d\theta[/tex]

[tex]= \frac{a^2}{\pi a^2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} d\theta = \frac{a^2}{\pi a^2} \left( \theta \Biggr|_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{a^2}{\pi a^2} \left(\frac{\pi}{2} + \frac{\pi}{2}\right) = 1[/tex]

Evaluating the integral without converting back to x, for ##a>0## I get +1, as above. For ##a<0##, I get -1 (because the endpoints would be the other way up).

The ##a## constants cancelled, but ##a## impacted on the sign of the final value. I'm guessing this is necessary to consider when stating the limit: either a right-hand limit or a left-hand limit.

I want to convert back to x so that someone could check my reasoning if that's okay.

I get [itex]\theta = \arctan{\frac{x}{a}}[/itex].

Here then, I would have to apply the endpoints for x, that is ##\infty## and ##-\infty##. To do this, I believe limits are applied to constants for the endpoints: the upper bound is ##\beta##, the lower bound is ##\alpha##, and:

[tex]\frac{a^2}{\pi a^2} \left(\lim_{\beta \rightarrow \infty} \left(\arctan{\frac{\beta}{a}}\right) - \lim_{\alpha \rightarrow -\infty} \left(\arctan{\frac{\alpha}{a}}\right)\right)[/tex]

Again, this gives +1 if ##a>0##, and -1 if ##a<0##, since arctan(x/a) changes sign if ##a## changes sign.

I'm guessing it is at this point that I can re-introduce the limit for ##a \rightarrow 0^{\pm}##. As ##a \rightarrow 0^{\pm}##, ##\frac{1}{a} \rightarrow \pm\infty##; For ##\frac{x}{a}##, both ##x## and ##a## want to take the value to either positive or negative infinity. For ##a \rightarrow 0^{+}##, ##\frac{\beta}{a} \rightarrow \infty## and ##\frac{\alpha}{a} \rightarrow -\infty##; and for ##a \rightarrow 0^{-}##, ##\frac{\beta}{a} \rightarrow -\infty## and ##\frac{\alpha}{a} \rightarrow \infty##. Therefore:

[tex]\lim_{a \rightarrow 0^{+}} \left( \frac{a^2}{\pi a^2} \left(\lim_{\beta \rightarrow \infty} \left(\arctan{\frac{\beta}{a}}\right) - \lim_{\alpha \rightarrow -\infty} \left(\arctan{\frac{\alpha}{a}}\right)\right) \right) = \frac{a^2}{\pi a^2} \left(\frac{\pi}{2} - \frac{-\pi}{2}\right) = \frac{a^2}{\pi a^2} \left(\frac{\pi}{2} + \frac{\pi}{2}\right) = 1[/tex]

or

[tex]\lim_{a \rightarrow 0^{-}} \left( \frac{a^2}{\pi a^2} \left(\lim_{\beta \rightarrow \infty} \left(\arctan{\frac{\beta}{a}}\right) - \lim_{\alpha \rightarrow -\infty} \left(\arctan{\frac{\alpha}{a}}\right)\right) \right) = \frac{a^2}{\pi a^2} \left(\frac{-\pi}{2} - \frac{\pi}{2}\right) = -1[/tex]

Okay, in both cases - either substituting back or not - I get the same results. However, as mentioned earlier, the limit is different from both sides for the limit of ##a##; this means I would have to explicitly state the direction for it to be meaningful, but the limit still exists. Is this in any way accurate?

The other question concerns an alternative solution to the integral. Wolfram Alpha churns out:

[tex]a \sqrt{\frac{1}{a^2}}[/tex]

It's also possible to see from this how the direction of the limit for ##a## affects the value, being 1 or -1. Does anyone know how wolfram got this solution?

Any help appreciated.

I have a couple of questions on the following. Firstly, I was hoping someone could check my working and my reasoning.

Secondly, I was wondering if someone knew an alternative way of solving this problem.

I wanted to integrate this from [itex]x = -\infty[/itex] to [itex]x = \infty[/itex]:

[tex]\lim_{a \rightarrow 0^{\pm}} \left(\frac{a}{\pi(x^2 + a^2)}\right)[/tex]

It looks kind of like the standard Cauchy distribution-thingy...

I figured since [itex]a \neq 0[/itex] then I can ignore the limit (##a## could just be a really small number, for example, or a large number, and, as a constant, except for the sign it wouldn't make a difference what its value is). I put [itex]a \rightarrow 0^{\pm}[/itex] rather than just ##a \rightarrow 0## because the limit is different from either side, which is also important. I think...

Therefore:

[tex]\int_{-\infty}^{\infty} \left( \lim_{a \rightarrow 0^{\pm}} \left(\frac{a}{\pi(x^2 + a^2)}\right) \right) dx = \lim_{a \rightarrow 0^{\pm}} \left( \int_{-\infty}^{\infty}\frac{a}{\pi(x^2 + a^2)} dx \right)[/tex]

I used a trig substitution [itex]x = a \ tan(\theta)[/itex] since the range of ##\theta## is the same as x: [itex]-\infty < \tan(\theta) < \infty[/itex]. To change the endpoints to ##\theta##, for a>0 I get the lower bound to be [itex]\frac{-\pi}{2}[/itex] and the upper bound to be [itex]\frac{\pi}{2}[/itex], but for ##a<0## I get ##\frac{\pi}{2}## and ##\frac{-\pi}{2}## (they 'flip'); for the example, using ##a>0##:

[tex]\int_{-\infty}^{\infty}\frac{a}{(x^2 + a^2)} dx = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{a}{\pi((a \tan(\theta))^2 + a^2)}\,a (\tan^2(\theta) + 1) d\theta = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{a^2 (\tan^2(\theta) + 1)}{\pi a^2 (\tan^2(\theta) + 1)} d\theta[/tex]

[tex]= \frac{a^2}{\pi a^2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} d\theta = \frac{a^2}{\pi a^2} \left( \theta \Biggr|_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{a^2}{\pi a^2} \left(\frac{\pi}{2} + \frac{\pi}{2}\right) = 1[/tex]

Evaluating the integral without converting back to x, for ##a>0## I get +1, as above. For ##a<0##, I get -1 (because the endpoints would be the other way up).

The ##a## constants cancelled, but ##a## impacted on the sign of the final value. I'm guessing this is necessary to consider when stating the limit: either a right-hand limit or a left-hand limit.

I want to convert back to x so that someone could check my reasoning if that's okay.

I get [itex]\theta = \arctan{\frac{x}{a}}[/itex].

Here then, I would have to apply the endpoints for x, that is ##\infty## and ##-\infty##. To do this, I believe limits are applied to constants for the endpoints: the upper bound is ##\beta##, the lower bound is ##\alpha##, and:

[tex]\frac{a^2}{\pi a^2} \left(\lim_{\beta \rightarrow \infty} \left(\arctan{\frac{\beta}{a}}\right) - \lim_{\alpha \rightarrow -\infty} \left(\arctan{\frac{\alpha}{a}}\right)\right)[/tex]

Again, this gives +1 if ##a>0##, and -1 if ##a<0##, since arctan(x/a) changes sign if ##a## changes sign.

I'm guessing it is at this point that I can re-introduce the limit for ##a \rightarrow 0^{\pm}##. As ##a \rightarrow 0^{\pm}##, ##\frac{1}{a} \rightarrow \pm\infty##; For ##\frac{x}{a}##, both ##x## and ##a## want to take the value to either positive or negative infinity. For ##a \rightarrow 0^{+}##, ##\frac{\beta}{a} \rightarrow \infty## and ##\frac{\alpha}{a} \rightarrow -\infty##; and for ##a \rightarrow 0^{-}##, ##\frac{\beta}{a} \rightarrow -\infty## and ##\frac{\alpha}{a} \rightarrow \infty##. Therefore:

[tex]\lim_{a \rightarrow 0^{+}} \left( \frac{a^2}{\pi a^2} \left(\lim_{\beta \rightarrow \infty} \left(\arctan{\frac{\beta}{a}}\right) - \lim_{\alpha \rightarrow -\infty} \left(\arctan{\frac{\alpha}{a}}\right)\right) \right) = \frac{a^2}{\pi a^2} \left(\frac{\pi}{2} - \frac{-\pi}{2}\right) = \frac{a^2}{\pi a^2} \left(\frac{\pi}{2} + \frac{\pi}{2}\right) = 1[/tex]

or

[tex]\lim_{a \rightarrow 0^{-}} \left( \frac{a^2}{\pi a^2} \left(\lim_{\beta \rightarrow \infty} \left(\arctan{\frac{\beta}{a}}\right) - \lim_{\alpha \rightarrow -\infty} \left(\arctan{\frac{\alpha}{a}}\right)\right) \right) = \frac{a^2}{\pi a^2} \left(\frac{-\pi}{2} - \frac{\pi}{2}\right) = -1[/tex]

Okay, in both cases - either substituting back or not - I get the same results. However, as mentioned earlier, the limit is different from both sides for the limit of ##a##; this means I would have to explicitly state the direction for it to be meaningful, but the limit still exists. Is this in any way accurate?

The other question concerns an alternative solution to the integral. Wolfram Alpha churns out:

[tex]a \sqrt{\frac{1}{a^2}}[/tex]

It's also possible to see from this how the direction of the limit for ##a## affects the value, being 1 or -1. Does anyone know how wolfram got this solution?

Any help appreciated.

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