Definite Integration Test - 4

If we define $F_b(a) \; = \; \int_0^\infty \frac{e^{-ax} \sin bx}{x}\,dx$ for $a,b > 0$ then $F_b(a) \to 0$ as $a \to \infty$ , while $F'_b(a) \; = \; -\int_0^\infty e^{-ax} \sin bx\,dx \; = \; -\frac{b}{a^2 + b^2}$ and hence $F_b(a) \; = \; \tfrac12\pi - \tan^{-1}\tfrac{a}{b}$ Since $\sin3x = 3\sin x - 4\sin^3x$ we deduce that $\int_0^\infty \frac{e^{-ax} \sin^3(4x)}{x}\,dx \; = \; \tfrac34F_4(a) - \tfrac14F_{12}(a) \; = \; \tfrac14\pi - \tfrac34\tan^{-1}\tfrac{a}{4} + \tfrac14\tan^{-1}\tfrac{a}{12}$ making the desired integral $\tfrac14\pi - \tfrac34\tan^{-1}\tfrac54 + \tfrac14\tan^{-1}\tfrac{5}{12}$ and hence the answer is $1 + 4 + 3 + 5 = \boxed{13}$ .

A solution not as Brilliant as Mr @Mark Hennings one, but still:

$\large \displaystyle I = \int_0^{\infty} \frac{ e^{-5x} \sin^3(4x)}{x} dx$

Since $\sin(3x) = 3 \sin(x) - 4 \sin^3(x)$ , we can deduce that $\sin(12x) = 3 \sin(4x) - 4 \sin^3(4x)$ , or, $\sin^3(4x) = \frac{3 \sin(4x) - \sin(12x)}{4}$

$\large \displaystyle I = \frac34 \int_0^{\infty} \frac{ e^{-5x} \sin(4x)}{x} dx - \frac14 \int_0^{\infty} \frac{ e^{-5x} \sin(12x)}{x} dx$

From the Maclaurin series of the sine:

$\large \displaystyle I = \frac34 \int_0^{\infty} \frac{ e^{-5x}}{x} \sum_{n=0}^{\infty} \frac{(-1)^n (4x)^{2n+1}}{(2n+1)!} dx - \frac14 \int_0^{\infty} \frac{ e^{-5x} }{x} \sum_{n=0}^{\infty} \frac{(-1)^n (12x)^{2n+1}}{(2n+1)!} dx$

$\large \displaystyle I = \frac34 \sum_{n=0}^{\infty} \frac{ (-1)^n 4^{2n+1} }{(2n+1)!} \int_0^{\infty} \ e^{-5x} x^{2n} dx - \frac14 \sum_{n=0}^{\infty} \frac{ (-1)^n 12^{2n+1} }{(2n+1)!} \int_0^{\infty} e^{-5x} x^{2n} dx$

On both integrals, make $x = \frac{t}{5}$ , $dx = \frac{dt}{5}$ :

$\large \displaystyle I = \frac34 \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1)!} \left (\frac{4}{5} \right )^{2n+1} \int_0^{\infty} \ e^{-t} t^{2n} dt - \frac14 \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1)!} \left (\frac{12}{5} \right )^{2n+1} \int_0^{\infty} e^{-t} t^{2n} dt$

Both integrals will be equal to $\Gamma(2n+1) = (2n)!$ ( $\Gamma(z)$ is the Gamma Function ). Also, since $(2n+1)! = (2n)! \cdot (2n+1)$ , one has:

$\large \displaystyle I = \frac34 \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1)} \left (\frac{4}{5} \right )^{2n+1}- \frac14 \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1)} \left (\frac{12}{5} \right )^{2n+1}$

From the Maclaurin series of the arc-tangent:

$\large \displaystyle I = \frac34 \arctan \left ( \frac{4}{5} \right ) - \frac14 \arctan \left ( \frac{12}{5} \right )$

Since $\arctan(x) = \frac{\pi}{2} - \arctan \left ( \frac{1}{x} \right )$

$\large \displaystyle I = \frac34 \left [ \frac{\pi}{2} - \arctan \left ( \frac{5}{4} \right ) \right ] - \frac14 \left [ \frac{\pi}{2} - \arctan \left ( \frac{5}{12} \right ) \right]$

$\color{#20A900} \boxed{ \large \displaystyle I = \frac{\pi}{4} - \frac34 \arctan \left ( \frac{5}{4} \right ) + \frac14 \arctan \left ( \frac{5}{12} \right ) }$

So:

$\color{#3D99F6} \large \displaystyle a =1, b = 4, c = 3, d = 5 \rightarrow \boxed{ \large \displaystyle a+b+c+d = 13}$