Hard Integration

Relevant wiki: Differentiation Under the Integral Sign

Similar solution with @Kelvin Hong's using differentiation under the integral sign.

$\begin{aligned} I(a) & = \int_0^\infty \frac {e^{-ax}\sin x}x dx & \small \color{#3D99F6} \text{Differentiate w.r.t. }a. \\ \frac {\partial I(a)}{\partial a} & = - \int_0^\infty e^{-ax} \sin x \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = e^{-ax} \cos x \bigg|_0^\infty + a \int_0^\infty e^{-ax} \cos x \ dx \\ & = 0-1 + a e^{-ax} \sin x \bigg|_0^\infty + a^2 \color{#3D99F6} \int_0^\infty e^{-ax} \sin x \ dx & \small \color{#3D99F6} \text{Note that }\frac {\partial I(a)}{\partial a} = - \int_0^\infty e^{-ax} \sin x \ dx \\ & = -1 + 0-0 {\color{#3D99F6}-} a^2 \cdot \color{#3D99F6} \frac {\partial I(a)}{\partial a} \\ & = - \frac 1{a^2+1} \\ \implies I(a) & = - \int \frac 1{a^2+1} da \\ & = - \arctan a +\color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ I(\infty) & = - \arctan \infty + C = 0 & \small \color{#3D99F6} \text{Note that }I(\infty) = 0 \\ & = - \frac \pi 2 + C = 0 & \small \color{#3D99F6} \implies C = \frac \pi 2 \\ \implies I(2) & = \frac \pi 2 - \arctan 2 & \small \color{#3D99F6} \text{Note that } I(2) \text{ is the required integral.} \end{aligned}$

Therefore, $a+b = 2+2 = \boxed 4$ .

Relevant Wiki: Differentiation Under the Integral Sign

Let $\displaystyle f(t)=\int_0^\infty\dfrac{e^{-tx}\sin x}{x}dx,$ by Differentiation under the integral sign: $\begin{aligned}f'(t)&=\int_0^\infty\dfrac{\partial}{\partial t}\dfrac{e^{-tx}\sin x}{x}dx\\&=\int_0^\infty-e^{-tx}\sin xdx\end{aligned}$

By using integration by part, we have the table below:

$\begin{array}{c|c}\sin x&-e^{-tx}\\\cos x&\frac1te^{-tx}\\-\sin x&-\dfrac1{t^2}e^{-tx}\end{array}$

So we ended up with

$\begin{aligned} f'(t)&=\dfrac{e^{-tx}\sin x}{t}+\dfrac{e^{-tx}\cos x}{t^2}-\dfrac1{t^2}f'(t)\\ (1+t^2)f'(t)&=e^{-tx}(t\sin x+\cos x)\\ f'(t)&=\dfrac{t\sin x+\cos x}{1+t^2}e^{-tx}\Bigg|_0^\infty\\ &=-\dfrac{1}{1+t^2}\\ \therefore f(t)&=-\arctan t+C \end{aligned}$

To calculate $C$ , we may consider $f(\infty)$ , so

$\displaystyle f(\infty)=\lim_{t\to\infty}\int_0^\infty\dfrac{e^{-tx}\sin x}{x}dx=0$ which help us to determine that $C=\dfrac{\pi}2$ Compute $t=2$ , we now have $\begin{aligned}\int_0^\infty{\dfrac{e^{-2x}\sin x}{x}}dx&=f(2)\\&=\frac{\pi}{2}-\arctan 2\end{aligned}$

$\int_0^{\infty}\frac{e^{-2x}Sinx}{x}dx=\frac {\pi}{a}- arctan(b)$ This integration can be evaluated by exploiting Feynman's trick. Considering the parameterisation $\implies I(a)=\int_0^{\infty}\frac{Sinx}{x}e^{-ax}dx$ $a\geq 0$ $\implies I(a)=-\int_0^{\infty}e^{-ax}Sinxdx$

Differentiating w.r.t $x$ $\implies I'(a)=-\frac{1}{1+a^2}$ Note that, since we are interested in $I(2)$ and we got $I(0)=\frac{\pi}{2}$

We obtain that, $-\int_0^2\frac {1}{1+a^2}da=-[I(2)-I(0)]=-[arctan (2)\frac {\pi}{2}]$ $\implies\frac {\pi}{2}-arctan (2)=arctan (\frac {1}{2})$

The last equality follows from the relationship between the inverse tangent and inverse co-tangent function. The value of $a+b=4$