I am circling in rounds by evaluating the following integral: $$J=\int_{0}^{T} e^{-xt} \cos t\, dt$$ where $x \to \infty$ and $T>0$
I started by integrating by parts. $$\int_{0}^{T} e^{-xt} \cos t\, dt = \bigg[\frac{-1}{x}e^{-xt}\cos t \bigg]_0^T- \int_{0}^{T} \frac{1}{x}e^{-xt} \sin t \,dt$$ But in stead of easier it became harder. I don't think a substitution is possible, but perhaps I am wrong.
Can anybody give me some help?
On
$J=\int_{0}^{T} e^{-xt} \cos (t) \,dt$
integrating by parts gives;
$J =e^{-xt}\cdot\sin(t)\bigg|_0^T+\int_0^T\frac{e^{-xt}}{x}\,\cos(t)\,dt $
integrate $\int_0^T\frac{e^{-xt}}{x}\,\cos(t)\,dt$ by parts once again;
$J = e^{-xt}\cdot\sin(t)\bigg|_0^T-\frac{e^{-xt}}{x}\,\cos(t)\bigg|_0^T-\int_0^T\frac{e^{-xt}}{x^2}\,\cos(t)\,dt$
$J =e^{-xt}\cdot\sin(t)\bigg|_0^T-\frac{e^{-xt}}{x}\,\cos(t)\bigg|_0^T-\frac{1}{x^2}J$
$J\bigg[1+\frac1{x^2}\bigg] = e^{-xT}\cdot\sin(T)-\frac{e^{-xT}}{x}\cdot\cos(T)+\frac1x$
$J =\dfrac{ e^{-xT}\cdot\sin(T)-\frac{e^{-xT}}{x}\cdot\cos(T)+\frac1x}{1+\frac1{x^2}}$
NOTE:
I am going to assume you meant $T\to \infty$, and proceed further,
$J = \lim_{T\to\infty}\bigg[\dfrac{ e^{-xT}\cdot\sin(T)-\frac{e^{-xT}}{x}\cdot\cos(T)+\frac1x}{1+\frac1{x^2}}\bigg]$
$J = \bigg[\dfrac{\frac{1}{x}}{\frac{x^2+1}{x^2}}\bigg]$
$J = \dfrac x{x^2+1}$
On
I don't understand your $x\to\infty$ under the integral notation. Maybe you mean $\lim_{x\to\infty}$ after computing the integral?
One can solve these type of integrals using complex numbers, because of $e^{iy}=\cos(y)+i\sin(y)$.
Define $$J' =\int_{0}^{T} e^{-xt} \sin t\, dt$$ so that
$$J+iJ' = \int_{0}^{T} e^{-xt}e^{it} \, dt = \int_{0}^{T} e^{(-x+i)t} \, dt = \frac{e^{(-x+i)T}-1}{-x+i}$$ and now extract $$J = \operatorname{Re}\bigg(\frac{e^{(-x+i)T}-1}{-x+i}\bigg)= \frac{x}{1+x^2} - \frac{e^{-Tx}x\cos(T)}{1+x^2}+\frac{e^{-Tx}\sin(x)}{1+x^2}$$
At least three methods spring to mind:
Using integration by parts, we have \begin{align} J &= \int_{0}^{T} \mathrm{e}^{-xt}\cos(t)\,\mathrm{d}t \\ &= \left.-\frac{1}{x}\cos(t)\mathrm{e}^{-xt}\right|_{t=0}^{T} - \frac{1}{x} \int_{0}^{T} \mathrm{e}^{-xt}\sin(t)\,\mathrm{d}t && \text{(IBP)} \\ &= -\frac{1}{x} \cos(T)\mathrm{e}^{-xT} + \frac{1}{x} - \left( \left. -\frac{1}{x^2} \sin(t) \mathrm{e}^{-xt} \right|_{t=0}^{T} + \frac{1}{x}^2 \int_{0}^{T} \mathrm{e}^{-xt} \cos(t) \,\mathrm{d}t \right) &&\text{(IBP again)} \\ &= -\frac{1}{x} \cos(T)\mathrm{e}^{-xT} + \frac{1}{x} + \frac{1}{x^2}\sin(T)\mathrm{e}^{-xT} - \frac{1}{x^2} J. \end{align} Solving for $J$, we obtain \begin{align} &\left( 1 + \frac{1}{x^2} \right) J = \left( \frac{x^2 + 1}{x^2} \right) = -\frac{1}{x} \cos(T)\mathrm{e}^{-xT} + \frac{1}{x} + \frac{1}{x^2}\sin(T)\mathrm{e}^{-xT} \\ &\qquad\implies J = -\frac{x}{x^2+1}\cos(T)\mathrm{e}^{-xT} + \frac{x}{x^2+1} + \frac{1}{x^2 + 1} \sin(T)\mathrm{e}^{-xT}. \end{align} Since both $\frac{1}{x}$ and $\mathrm{e}^{-xT}$ go to zero as $x \to \infty$, it follows that $\lim_{x\to\infty} J = 0$.
The downside to this approach, I think, is that there is a lot of computation, which means that there are a lot of places where we can make mistakes. Indeed, I am not entirely certain that all of my computations are correct (I found two mistakes while typing my handwritten notes into the answer box—it seems reasonable to conclude that there are other mistakes still remaining). I'm sure that the basic order estimates are correct, but the exact formula for $J$ scares me a bit.
We can use the fact that $$ \cos(t) = \frac{\mathrm{e^{it} + \mathrm{e}^{-it}}}{2} $$ and integrate. Via this approach, we have \begin{align} J &= \int_{0}^{T} \mathrm{e}^{-xT}\cos(t) \,\mathrm{d}t \\ &= \frac{1}{2} \int_{0}^{T} \mathrm{e}^{-xt} \left( \mathrm{e}^{it} + \mathrm{e}^{-it} \right)\,\mathrm{d}t \\ &= \frac{1}{2} \int_{0}^{T} \mathrm{e}^{(-x+i)t} + \mathrm{e}^{(-x-i)t}\,\mathrm{d}t \\ &= \frac{1}{2} \left[ \frac{1}{-x+i} \mathrm{e}^{(-x+i)t} + \frac{1}{-x-i} \mathrm{e}^{(-x-i)t} \right|_{t=0}^{T} \\ &= \frac{1}{2} \left[ \frac{1}{i-x} \mathrm{e}^{(i-x)T} - \frac{1}{x+i} \mathrm{e}^{-(x+i) T} - \frac{1}{i-x} + \frac{1}{x+i} \right]. \end{align} Again, taking $x\to \infty$, this goes to zero. Unfortunately, there is still a great deal of computation, and it is easy to make mistakes, though we do avoid two integration by parts steps, which makes me more confident about the result. Note that the exact formula for $J$ looks different, but after simplification, I'm sure that (in particular) all of the $i$s will fall out and we'll get something recognizable.
I think that the best approach to this problem is to apply the Dominated Convergence Theorem. On the interval $[0,T]$, the function $$ f(t) = \mathrm{e}^{-xt} \cos(t) $$ is bounded by $1$ for any $x > 0$. That is, $$ |f(t)| = \left| \mathrm{e}^{-xt} \cos(t) \right| = \left| \mathrm{e}^{-xt} \right| |\cos(t)| \le \left| \mathrm{e}^{0} \right| = 1. $$ Since $\int_{0}^{T} 1 \,\mathrm{d}t = T < \infty$, we can apply the Dominated Convergence Theorem in order to conclude that $$ \lim_{n\to \infty} \int_{0}^{T} \mathrm{e}^{-nt}\cos(t) \,\mathrm{d}t = \int_{0}^{T} \lim_{n\to\infty} \left( \mathrm{e}^{-nt}\cos(t) \right) \,\mathrm{d}t, $$ where $n$ is a positive integer. But $$ \lim_{n\to\infty} \left( \mathrm{e}^{-nt}\cos(t) \right) = 0 $$ for all $t \in [0,T]$, so we have $$ \lim_{n\to\infty} J = \lim_{n\to \infty} \int_{0}^{T} \mathrm{e}^{-nt}\cos(t) \,\mathrm{d}t = \int_{0}^{T} 0 \,\mathrm{d}t = 0. $$ This approach is nice in that there is very little computation, but there is a detail that I have elided, namely that I have only proved the result for $n\in\mathbb{N}$, not for $x\in[0,\infty)$. However, the argument can be modified to make it rigorous with out too much difficulty.