Question

I'm putting my work in the comments as a picture. Someone helped me get this far but I don't quite understand why we did a few things and I don't understand the evaluation. Could someone look it over and let me know what you think? Thanks!!

For what values of a is \[\int\limits_{0}^{\infty}e ^{ax}\cos x dx\] convergent? Evaluate the integral for those values of a.

Answer 1

Someone helped me with this and this is what I have but they didn't think it was correct and there are parts I just don't understand.....
One of them is - I don't understand why we did the 2nd integration by parts. It appears that we ended up with an equation similar to what we had after the 1st integration by parts.....then I don't understand the ending with the evaluations....
(If this is even correct)

Answer 2

its a by parts method

Answer 3

when you start finding yourself going in circles, theres a way out :)

Answer 4

notice for simplicity:\[A = n - kA\]
this represents going in a circle, the left and right side both have A in them ... do some algebra.
\[A+kA = n\]
\[A(1+k) = n\]
\[A = \frac{n}{1+k}\]
the algebra breaks the circle and gives us solid result

Answer 5

your question is similar, and is prolly trying to get you familiar with, something that will be called a LaPlace transform later on:
\[L\{f(t)\}(s)=\int_{0}^{\infty}~e^{-st}~f(t)~dt\]

Answer 6

\[\begin{matrix}
\pm&&e^{ax}
\\+&f(x)&\frac1ae^{ax}
\\-&f'(x)&\frac1{a^2}e^{ax}
\\+&f''(x)&\frac1{a^3}e^{ax}\\
\\...&...&...
\end{matrix}\]

this gives us the setup:\[e^{a\infty}\left(\frac{f(\infty)}{a}-\frac{f'(\infty)}{a^2}+\frac{f''(\infty)}{a^3}\pm...\right)-e^{0}\left(\frac{f(0)}{a}-\frac{f'(0)}{a^2}+\frac{f''(0)}{a^3}\pm...\right)\]
\[e^{a\infty}\left(\frac{f(\infty)}{a}-\frac{f'(\infty)}{a^2}+\frac{f''(\infty)}{a^3}\pm...\right)-\left(\frac{f(0)}{a}-\frac{f'(0)}{a^2}+\frac{f''(0)}{a^3}\pm...\right)\]
this has the only hope of converging only if as \(x\to \infty\) then \(e^{(a\infty)}\to 0\).

this happens at e^(-00), so let a = -s

\[e^{-s\infty}\left(-\frac{f(\infty)}{-s}\frac{f'(\infty)}{s^2}-\frac{f''(\infty)}{s^3}-...\right)-\left(-\frac{f(0)}{s}-\frac{f'(0)}{s^2}-\frac{f''(0)}{s^3}-...\right)\]
\[0+\frac{f(0)}{s}+\frac{f'(0)}{s^2}+\frac{f''(0)}{s^3}+...\]

Answer 7

to work on the cos(x) we take successive derivativies:

cos
-sin
-cos
sin
cos ... back at the same place ...so its cyclic

cos0 = 1
-sin0 = 0
-cos0 = -1
sin 0 = 0

every other f(0) is 0, and the signs alternate, given us:

\[\frac{1}{s}+\frac{0}{s^2}-\frac{1}{s^3}+\frac{0}{s^4}+\frac{1}{s^5}+...\]
\[\frac{1}{s}-\frac{1}{s^3}+\frac{1}{s^5}-...\]
\[\sum_{n=0}^{\infty}\frac{(-1)^n}{s^{2n+1}}\]

Answer 8

Thank you!