$\large \int_0^\infty \sin{x^{5}} \ dx = \frac 1a \left(\sqrt b-1\right) \Gamma \left(\frac 1c \right)$

The equation above holds true for positive integers $a$ , $b$ , and $c$ . Submit $abc$ .

**
Notation:
**
$\Gamma(\cdot)$
denotes the
gamma function
.

**
Hint:
**
Use
$e^{ix}= \cos{x} + i\sin{x}$
.

The answer is 500.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

$\begin{aligned} I & = \int_0^\infty \sin x^5 \ dx \\ & = \int_0^\infty \frac {i \left(e^{-ix^5} - e^{ix^5}\right)}2 \ dx \\ & = \frac i2 {\color{#3D99F6} \int_0^\infty e^{-ix^5} dx} - \frac i2 {\color{#D61F06} \int_0^\infty e^{ix^5} dx} & \small \color{#3D99F6} u = ix^5 \implies du = 5ix^4 dx \\ & = \frac {\color{#3D99F6}e^{-\frac {2\pi i}5}}{10} {\color{#3D99F6} \int_0^\infty u^{-\frac 45} e^{-u} du} {\color{#D61F06} +} \frac {\color{#D61F06}e^{\frac {2\pi i}5}}{10} {\color{#3D99F6} \int_0^\infty u^{-\frac 45} e^{-u} du} & \small \color{#D61F06} u = -ix^5 \implies du = -5ix^4 dx \\ & = \frac {\cos \frac {2\pi}5}5 \Gamma \left(\frac 15 \right) \\ & = \frac 1{20}\left(\sqrt 5-1\right) \Gamma \left(\frac 15 \right) \end{aligned}$

$\implies abc = 20\times 5 \times 5 = \boxed{500}$