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I am taking single-variable calculus as part of my first year undergraduate mathematics course. I would like to compute the improper integral :

$$ \int_0^\infty e^{-ax}\cos {bx} \, dx $$

There is a brilliant answer to this question, using the techniques of complex analysis here.

However, not having taken a sequence in real and complex analysis yet, I am attempting to solve this using integration by parts. I am posting my solution here. The difficulty I am facing is in evaluating the limits.

Solution.

By the definition of an improper integral,

$$ I_c = \lim_{C\to\infty} \int_0^C e^{-ax}\cos bx \, dx $$

Let $u=e^{-ax}$ and $dv=\cos bx dx$.

Then, $du=-ae^{-ax}$ and $v=\frac{1}{b}\sin bx.$

Consequently, we have :

$$\begin{aligned} I_c&=\lim_{C\to\infty}\left[\frac 1 b e^{-ax} \sin bx\right]_0^C + \frac a b \lim_{C\to\infty}\int_0^C e^{-ax} \sin bx \, dx\\ &=\lim_{C\to\infty}\left[\frac{1}{b}e^{-ax} \sin bx\right]_0^C + \frac a b I_s \end{aligned}$$

We apply integration by parts again to $I_{s}$.

Let $u=e^{-ax}$ and $dv=\sin {bx}$.

Then, $du=-ae^{-ax}$ and $v=-\frac{1}{b}\cos bx$.

Consequently, we have :

$$\begin{aligned} I_s&=\lim_{C\to\infty}\left[-\frac{1}{b}e^{-ax} \cos bx\right]_0^C - \frac a b \lim_{C\to\infty} \int_0^C e^{-ax} \cos bx \, dx\\ &=\lim_{C\to\infty}\left[-\frac{1}{b}e^{-ax}\cos bx\right]_0^C - \frac a b I_c \\ \frac a b I_s&=\lim_{C\to\infty}\left[-\frac{a}{b^2}e^{-ax}\cos bx \right]_0^C-\frac{a^2}{b^2} I_c \end{aligned}$$

Substituting this result in the expression for $I_c$, we have:

$$\begin{aligned} I_c &=\lim_{C\to\infty}\left[\frac{1}{b}e^{-ax} \sin bx\right]_0^C-\lim_{C\to\infty}\left[\frac{a}{b^2}e^{-ax}\cos bx\right]_0^C-\frac{a^2}{b^2} I_c\\ \left(1+\frac{a^2}{b^2}I_c\right)&=\lim_{C\to\infty}\left[\frac{1}{b} e^{-ax} \sin bx\right]_0^C-\lim_{C\to\infty}\left[\frac{a}{b^2}e^{-ax}\cos bx\right]_0^C\\ \left(\frac{b^2+a^2}{b^2}\right)I_c&=\lim_{C\to\infty}\left[\frac{e^{-ax}(b\sin bx-a \cos bx)}{b^2}\right]_0^C \\ I_c &=\lim_{C\to\infty}\left[\frac{e^{-ax}(b\sin bx-a \cos bx)}{b^2+a^2} \right]_0^C \\ &=\frac 1 {b^2+a^2}\lim_{C\to\infty}\left[\frac{(b\sin bC-a \cos bC)}{e^{aC}}+a\right] \end{aligned}$$

I am not able to evaluate this limit -

$$ \lim_{C\to\infty}\left[\frac{(b\sin bC-a \cos bC)}{e^{aC}}\right] $$

I understand that, the numerator must oscillate and the denominator becomes infinitely large as $C\to\infty$. Can I therefore reason, that it approaches zero? Or is there any way, I can evaluate this limit?

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Use the property that the numerator is bounded and the denominator tend to infinity if $a>0$.

$$ \lim_{C\to{\infty}}\left|\frac{(b\sin {bC}-a \cos {bC})}{e^{aC}}\right|\le \lim_{C\to{\infty}}\frac{|a|+|b|}{e^{aC}} = (|a|+|b|) \lim_{C \to \infty} \exp(-aC)=0 $$

provided that $a>0$.

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