# Exponent tends to zero

Hey Rajesh,

I assume you are refering to the additional materials concerning Normal Distribution in the Probability course.

For future enquiries it is highly advisable to mark the question, pointing out the section which causes you trouble.

That would be of great help to instructors and you would be able to receive answers at shorter notice.

Apart from that, it's wonderful to see keen interest in the derivations of the Normal Distribution formulas. Let's dive in:

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First off, Step 7 suggest splitting into two simple integrals. The first one is the famous Gaussian integral, of which there are

numerous beautiful solutions. You can search it in the Internet or use Wolfram Alpha (further details in the pdf under

Probabilities / Continuos Distributions, check on the resources). Your immediate question is about the second integral and

Step 8. We go about it step by step:

- We integrate from -inf to +inf t * e^(-t^2) dt. To solve this integral, type it in Wolfram Mathematica or, if you are familiar with solving

indefinite integrals, transfer t to dt to obtain integral from -inf to +inf of (1/2) * e^(-t^2) d(t^2) (because d(t^2)/dt = 2t). Next, you want

to have d(-t^2) instead of d(t^2), so you add a minus in front of the integral and write d(-t^2). Following is direct integration and

application of Newton-Leibniz's theorem. That's how you get the result at the end of Step 7.

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Now, you have e^(-t^2). When t tends to +infinity (t -> +inf), then t^2 also tends to +infinity (t^2 -> +inf) and consequently

-t^2 tends to -infinity (because of the negative sign). It's a limit of e^(-t^2) at t tends to infinity - because t^2 will always tend

to +infinity when t tends to either +infinity or -infinity, it is the case that the exponent tends to zero. This answers the question.

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Try writing this in Wolfram Mathematica (https://www.wolframalpha.com/) and verify for yourself.

`int_{-inf}^{inf} e^(-t^2) dt`

will get you the value of the Gaussian integral

`lim_{t to +inf} e^(-t^2)`

will get you zero, because `lim_{t to +inf} -t^2`

gives -infinity.

Similarly, `lim_{t to -inf} e^(-t^2)`

is again zero since `lim_{t to -inf} -t^2`

is again -inifinity.

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Hope that has helped.

Best regards,

A., The 365 Team