# How did the highlighted part become zero ?

Discussion in 'Math' started by Damien De Silva, Mar 13, 2018.

May 20, 2017
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2. ### wayneh Expert

Sep 9, 2010
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What’s the value of that term at the two limits?

3. ### Damien De Silva Thread Starter New Member

May 20, 2017
28
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zero to infinity

4. ### wayneh Expert

Sep 9, 2010
16,099
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Try taking the limit as t approaches infinity. I think you’ll find it’s zero.

And my integration skills are a little rusty but I don’t think you need that substitution trickery. It’s only making it more confusing.

Last edited: Mar 13, 2018
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5. ### Damien De Silva Thread Starter New Member

May 20, 2017
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Ok Thanks a lot. I get it now.

6. ### wayneh Expert

Sep 9, 2010
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Well don’t trust my hunch without checking it. It’s just a hunch.

7. ### WBahn Moderator

Mar 31, 2012
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Not what are the values of 't' at the two limits. What is the value of the expression being evaluated at those two values of 't'?

8. ### MrAl AAC Fanatic!

Jun 17, 2014
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Hi,

I think i might see the reason for this question.

Maybe because we see a "minus s" in the exponent, not a "plus s" which would be a bit more ambiguous. I think the minus s (-s and not just s which is positive) means that 's' itself must be positive, or at least the real part of 's' must be positive (so that it is negative in the exponent due to the minus sign present in the definition).
Thus we end up with zero rather than infinity or something we cant define.
In fact, if we dont assume that then we cant solve this without some doubt as to if we did it right or not.
There may be other ways to explain this too.

To illustrate just a little...
e^(-s)=e^(-(a+bj))
and for the limit a must be positive and b does not matter as usual, but more direct 's' must be positive.