Is it even Possible?

Calculus Level 3

∫ 0 ∞ sin ⁡ x x d x = ? \large \int_0^\infty \frac{\sin x}{x} \, dx = \, ?

π \pi 8 π 8\pi π 4 \frac{\pi}{4} π 2 \frac{\pi}{2} 4 π 4\pi 2 π 2\pi π 2 \pi^2

View wiki
Gaurav Agarwal by Gaurav Agarwal , 38, Australia

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.

1 solution

I = ∫ 0 ∞ sin ⁡ x x d x = ∫ 0 ∞ sin ⁡ x ( ∫ 0 ∞ e − x y d y ) d x = ∫ 0 ∞ ( ∫ 0 ∞ e − x y sin ⁡ x d x ) d y \displaystyle I= \int_{0}^{\infty} \frac{\sin x}{x}dx=\int_{0}^{\infty} \sin x (\int_{0}^{\infty} e^{-xy}dy)dx = \int_{0}^{\infty} (\int_{0}^{\infty} e^{-xy}\sin x dx)dy

Now We know by simple IBP that ∫ 0 ∞ e − a x sin ⁡ b x d x = b b 2 + a 2 \displaystyle \int_{0}^{\infty} e^{-ax}\sin bx dx = \frac{b}{b^2+a^2} so,

I = ∫ 0 ∞ d y y 2 + 1 = [ tan ⁡ − 1 y ] 0 ∞ = π 2 \displaystyle I=\int_{0}^{\infty} \frac{dy}{y^2+1} = [\tan^{-1} y]_{0}^{\infty}=\boxed{\frac{\pi}{2}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...