5 once again

Calculus Level 5

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

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.

1 solution

Chew-Seong Cheong
Sep 29, 2015

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

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

The substitution $iy = x^5$ needs to be justified by contour integration. It cannot simply be used.

You need the fact that the integrand has no poles in the first quadrant, and that it tends to zero as $|z| \to \infty$ .

Mark Hennings - 5 years, 4 months ago

Nice Solution. I too did it in the same way.

Surya Prakash - 5 years, 8 months ago

I think this question is overrated...Do you think so?

Kishore S. Shenoy - 5 years, 7 months ago

I am not particular on the rate. I think the system will resolve it when more members try it.

Chew-Seong Cheong - 5 years, 7 months ago

@Chew-Seong Cheong – Yeah, that's true...but for the present, I feel it is so...

Kishore S. Shenoy - 5 years, 7 months ago

@Chew-Seong Cheong Sir, can you please verify the answer of this question

Kishore S. Shenoy - 5 years, 7 months ago

If you make the substitution $iy=x^5$ , then the bounds of integration on $y$ become $y=0\rightarrow y=-i \infty$ . In other words, an integration along the positive real axis in $x$ corresponds to an integration along the negative imaginary axis in $y$ . Your solution does not reflect this.

5 once again

The answer is 500.

1 solution

0 pending reports