Solving the kidnapped integral

POV: someone thinks KIDNAPPED THE OTHER ∫₋∞^∞ sin x⁄x dx I(t) = ∫₀^∞ sin x⁄x e⁻ᵗˣ dx I′(t) = ∂/∂t ∫₀^∞ sin x⁄x e⁻ᵗˣ dx ℒ{sin(ωt)} = ω⁄(s² + ω²) limₜ→∞ (−arctan(t) + C) = −π⁄2 + C limₜ→∞ ∫₀^∞ sin(x)e⁻ᵗˣ dx = 0 I(0) = π⁄2 − arctan(0) = π⁄2 − 0 = π⁄2

A TikTok-style video begins with a kidnapped-person joke, then presents the impossible-looking improper integral ∫₋∞^∞ sin(x)/x dx. A blond presenter and a flamboyant fantasy character work through the solution using a parameterized integral, differentiation under the integral sign, and a Laplace transform, finally obtaining π/2; the clip ends on an angry reaction face. The overly serious mathematical rescue mission is paired with Luude’s energetic drum-and-bass remix of “Down Under” featuring Colin Hay, making the absurd joke feel dramatic and triumphant. EN: A TikTok-style video begins with a kidnapped-person joke, then presents the impossible-looking improper integral ∫₋∞^∞ sin(x)/x dx. A blond presenter and a flamboyant fantasy character work through the solution using a parameterized integral, differentiation under the integral sign, and a Laplace transform, finally obtaining π/2; the clip ends on an angry reaction face. The overly serious mathematical rescue mission is paired with Luude’s energetic drum-and-bass remix of “Down Under” featuring Colin Hay, making the absurd joke feel dramatic and triumphant.

As kidnapped. Agatha needs your help! You must solve this integral to reactivate Agatha's force field! Good luck, my child! Bro, what the fuck am I meant to do with this? I thought sine x over x had no elementary anti-derivative. But there is still a way to get a value for this improper integral. I see! We basically define a function that is equal to our integral at a certain point to turn it into an initial value problem. Using Feynman's trick, we can differentiate under the integral sine. So now we can just use integration by parts here? It's quicker to do the Laplace transform of sine. Oh yeah, so it'll be negative 1 over 1 plus t squared. Now what though? We need a value for c? All we gotta do now is compare the limits as t approaches infinity of both expressions of i of t yielding this simple equation. Alright, so we have the specific solution to the initial value problem. Now we just need to sub in zero. Marvelous work, my son! You have re-sheathed the Garthas defenses! You have saved your race! A Gotha

♪ Luude — Down Under (feat. Colin Hay)

• @tgkshitandpost / 12025 2025-12-09