Avant Garde

Consider the integral,
$\displaystyle I(a) = \int_0^{\infty } e^{-at} \cos t \text{ d}t$
$\displaystyle \Rightarrow - I'(a) = \int_0^{\infty} t e^{-at} \cos t \text{ d}t$ ,

which is the required value. Our whole effort is now towards finding out the value of $I(a)$ .

$\displaystyle \begin{array}{c}\\ I(a) && = \int_0^{\infty } e^{-at} \cos t \text{ d}t \\ && = \mathfrak{R} \left( \int_0^{\infty } e^{-at} e^{it} \text{ d}t \right) \\ && = \mathfrak{R} \left( \frac{e^{(-a+i)t}}{-a + i } |_0^{\infty} \right) \\ && = \mathfrak{R} \left( \frac{1}{a - i} \right) \\ && = \frac{1}{a^2 + 1} \\ \end{array}$

$\displaystyle \Rightarrow - I'(a) = \frac{a^2 -1}{(a^2 + 1)^2}$ .

We want to calculate $- I'(2)$ which evaluates to $\boxed{ \dfrac{3}{25}}$ .

Note: For those who know Laplace transforms it it pretty easy to realize that $I(a)$ is nothing but $\mathfrak{L} ( \cos t )$ which is a standard value.

The integral is equivalent to finding the Laplace Transform of $t\cdot cos(t)$ evaluated at $s=2$ . $Laplace(t\cdot cos(t))=\frac{s^2-1}{(s^2+1)^2}$ . Evaluating the transform at $s=2$ , gives you $\frac{3}{25}$ .

My solution is not like the other two. My solution goes like,

Let $I_{1}= \displaystyle{\int_{0}^{\infty} t e^{-2t} \cos t dt}$ and $I_{2}= \displaystyle{\int_{0}^{\infty} t e^{-2t} \sin t dt}$

Now If we take $I_{1}+i I_{2}$ where $i=\sqrt{-1}$ .

So we get $I_{1}+iI_{2}=\displaystyle{\int_{0}^{\infty} t e^{-2t} (\cos t+ i \sin t) dt}$

$\implies I_{1}+iI_{2}=\displaystyle{\int_{0}^{\infty} t e^{-2t} (e^{it}) dt}$

$\implies I_{1}+iI_{2}=\displaystyle{\int_{0}^{\infty} t(e^{it-2t}) dt}$

After solving the integral taking i to be just a constant(It is a constant :P anyway)!!

We get $I_{1}+iI_{2} = \dfrac{3+4i}{25}$

So equating real part, we get $I_{1} = \dfrac{3}{25} = \boxed{0.12}$ .

By the way, through my method, We also get another integral $I_{2}= \displaystyle{\int_{0}^{\infty} t e^{-2t} \sin t dt}= \dfrac{4}{25}$