How is


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 $\infty$


  • $\begingroup$ 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
  • 1
    $\begingroup$ 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


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:

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

  • $\begingroup$ 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

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 $T$:

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


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

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


which should also cause no trouble.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @UmerFarooq He's hinting one integral will converge, but the other will not. $\endgroup$ – Pedro Tamaroff Oct 31 '12 at 20:56

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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.