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Consider the limit $$ \lim_{n \to \infty} \int_0^n e^{-px} \cos(ax)\, dx.$$

The exercise is asking about the values of $p$ and $a$ which the limit above exists.

a) exists for infinity values of $p$

b) exists for a finite values of $a$

c) exists for every real $p$ but not for every value of $a$

d) exists for every real $p$ and real $a$

e) none of the above

After a while I'm guessing the answer d) is correct. But I am not sure. Also, asking if the limit above exists, is the same thing as asking if the integral converges? If I evaluate the limit for a negative $p$ and a positive $a$ the integral oscillates between -infinity and infinity. That means the integral diverges and the limit above doesn't exist? I will be thankful if someone can provide a answer and a explanation.

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  • $\begingroup$ What do "exists for infinity values of p" and "exists for a finite values of a" mean? $\endgroup$ – Antonio Vargas Sep 26 '17 at 7:26
  • $\begingroup$ @AntonioVargas "does the limit exists for all real number p?" or "the limit exists for a number A restricted in a interval? for example: A is natural and less than 10" $\endgroup$ – Pinteco Sep 26 '17 at 18:34
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Hint. Note that, by integration by parts, if $p^2+a^2\not=0$ then $$\int e^{-px} \cos(ax)\, dx=\frac{e^{-px}(a\sin(ax)-p\cos(ax))}{p^2+a^2}.$$ Is it easier to go on now? Does the limit exist for $p=-1$ and $a=\pi$?

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  • $\begingroup$ the limit oscillates between -infinity and infinity. Can I say that the limit exists but the integral diverges? $\endgroup$ – Pinteco Sep 26 '17 at 18:56
  • $\begingroup$ @Anon No, in that case the limit does not exist. It exists for $p>0$. $\endgroup$ – Robert Z Sep 26 '17 at 20:20
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Well, notice we are looking at the Laplace transform:

$$\mathscr{L}_x\left[\cos\left(\text{a}\cdot x\right)\right]_{\space\left(\text{p}\right)}:=\int_0^\infty\cos\left(\text{a}\cdot x\right)\cdot e^{-\text{p}t}\space\text{d}t=\frac{\text{p}}{\text{p}^2+\text{a}^2}\tag1$$

When $\Re\left(\text{p}\right)>0$

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  • $\begingroup$ I didn't know about this Laplace transform. What does R(p)>0 means? $\endgroup$ – Pinteco Sep 26 '17 at 18:58
  • $\begingroup$ @Anon It means that the real part of $\text{p}$ should be bigger than $0$ in order to have a finite value for the integral. $\endgroup$ – Jan Sep 27 '17 at 10:40

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