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How is

$$\lim_{T\to\infty}\frac{1}T\int_{-T/2}^{T/2}e^{-2at}dt=\infty\;?$$

however my answer comes zero because putting limit in the expression, we get:

$$\frac1\infty\left(-\frac1{2a}\right) [e^{-\infty} - e^\infty]$$ which results in zero?

I think I am doing wrong. So how can I get the answer equal to $\infty$

Regards

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  • $\begingroup$ what is Latex ? $\endgroup$ – Umer Farooq Oct 31 '12 at 20:31
  • $\begingroup$ yeah sure there was 1/T after limit $\endgroup$ – Umer Farooq Oct 31 '12 at 20:35
  • $\begingroup$ @BabakSorouh I've put it back in. $\endgroup$ – Thomas Andrews Oct 31 '12 at 20:36
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    $\begingroup$ This expression: $$\frac1\infty\left(-\frac1{2a}\right) [e^{-\infty} - e^\infty]$$ doesn't mean anything. How come you say it is zero? $\endgroup$ – Thomas Andrews Oct 31 '12 at 20:36
  • $\begingroup$ @Babak: Yep; I probably lost it somehow when I put in the limits of integration. $\endgroup$ – Brian M. Scott Oct 31 '12 at 20:36
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$I=\int_{-T/2}^{T/2}e^{-2at}dt=\left.\frac{-1}{2a}e^{-2at}\right|_{-T/2}^{T/2}=-\frac{e^{-aT}+e^{aT}}{2a}$

Despite that $a$ positive or negative, one of the exponents will tend to zero at our limit, so we can rewrite it as :

$\lim_{T\rightarrow\infty}\frac{I}{T}=\lim_{T\rightarrow\infty}\left(-\frac{e^{-aT}+e^{aT}}{2aT}\right)=\lim_{T\rightarrow\infty}\left(-\frac{e^{\left|a\right|T}}{2aT}\right) $

But becuase exponental functions are growing much faster than $T$ , this makes the limit is always infinity, thus finaly we have:

$\lim_{T\rightarrow\infty}\frac{I}{T}=\begin{cases} -\infty & a>0\\ +\infty & a<0 \end{cases}$

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  • $\begingroup$ how do you say that it is equal to ∞. Please explain a little $\endgroup$ – Umer Farooq Oct 31 '12 at 20:52
  • $\begingroup$ I added details in the answer. $\endgroup$ – TMS Oct 31 '12 at 21:01
  • $\begingroup$ I like your attempt TMS. $\endgroup$ – mrs Oct 31 '12 at 21:03
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The biggest thing that you’re doing wrong is trying to treat $\infty$ as if it were a number with which you can do arithmetic: it isn’t. You really do have to work with limits. Let’s start with the integral for some fixed value of $T$:

$$\begin{align*} \int_{-T/2}^{T/2}e^{-2at}dt&=-\frac1{2a}\left[e^{-2at}\right]_{-T/2}^{T/2}=-\frac1{2a}\left(e^{-aT}-e^{aT}\right)\\ &=-\frac1{2a}\left(\frac1{e^{aT}}-e^{aT}\right)\\ &=\frac1{2a}\left(e^{aT}-\frac1{e^{aT}}\right)\\ &=\frac{e^{2aT}-1}{2ae^{aT}}\;. \end{align*}$$

Thus,

$$\begin{align*} \lim_{T\to\infty}\frac{1}T\int_{-T/2}^{T/2}e^{-2at}dt&=\lim_{T\to\infty}\frac{e^{2aT}-1}{2aTe^{aT}}=\lim_{T\to\infty}\left(\frac{e^{2aT}}{2aTe^{aT}}-\frac1{2aTe^{aT}}\right)\\ &=\lim_{T\to\infty}\frac{e^{2aT}}{2aTe^{aT}}-\lim_{T\to\infty}\frac1{2aTe^{aT}}\;. \end{align*}$$

You should have no trouble evaluating $\lim_{T\to\infty}\dfrac1{2aTe^{aT}}$, and

$$\lim_{T\to\infty}\frac{e^{2aT}}{2aTe^{aT}}=\lim_{T\to\infty}\frac{e^{aT}}{2aT}\;,$$

which should also cause no trouble.

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I think you can split the integral into two as: $$\int_{-T/2}^{T/2}e^{-2at}dt=\int_{0}^{T/2}e^{-2at}dt-\int_{0}^{-T/2}e^{-2at}dt$$ and then use L'Hôpital's rule for the limit.

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  • $\begingroup$ N.T.: L'Hôpital or L'Hospital. $\endgroup$ – Pedro Tamaroff Oct 31 '12 at 20:46
  • $\begingroup$ no the integral wasn't split. If you put limits in the expression I have stated above, what answer do you get $\endgroup$ – Umer Farooq Oct 31 '12 at 20:51
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    $\begingroup$ @UmerFarooq He's hinting one integral will converge, but the other will not. $\endgroup$ – Pedro Tamaroff Oct 31 '12 at 20:56
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Assume wlog. $a>0$. By integration, $$\frac{1}{T}\int^{T/2}_{-T/2} e^{-2at} dt = \frac{1}{2aT}e^{aT} - \frac{1}{2aT}e^{-aT}$$

Note that the second summand goes to zero as $T\rightarrow\infty$ because $1/T$ and $e^{-aT}$ each go to zero.

However, the first summand goes to infinity, as can be seen most intuitively when looking at the Taylor expansion of the exponential function:

$$\frac{1}{T} e^{aT} = \frac{1}{T} \left( 1 + aT + \frac{(aT)^2}{2} + \cdots \right) = \frac{1}{T} + a + \frac{a^2T}{2} + \cdots$$

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