JEEIntermediate
System Entry: Jul 20, 2026Feynman Integration Trick
#calculus#integration
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Problem Statement
- Determine the structural parameter convergence bounds before evaluating the definite integral array.
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- \int_{0}^{\infty} \frac{\sin(x)}{x} , dx = \frac{\pi}{2}
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- Verify the result by defining a continuous parametric function $I(\alpha)$ and applying Leibniz’s integral rule for differentiation under the integral sign.
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Analytical Proof Structure
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1. Parametrization of the Kernel
- We introduce a parameter $\alpha \geq 0$ to construct an auxiliary function $I(\alpha)$:
- $$I(\alpha) = \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} , dx$$
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Boundary Conditions: Notice that our target integral is exactly $I(1)$, and the limiting behavior as the parameter approaches infinity yields $\lim_{\alpha \to \infty} I(\alpha) = 0$.
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2. Differentiation Under the Integral Sign
- Differentiating both sides with respect to $\alpha$ allows us to eliminate the problematic $x$ in the denominator:
- $$\frac{dI}{d\alpha} = \frac{d}{d\alpha} \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} , dx$$
- Assuming the conditions for Leibniz’s rule hold under uniform convergence, we pass the derivative inside:
- $$\frac{dI}{d\alpha} = \int_{0}^{\infty} \frac{\partial}{\partial \alpha} \left( \frac{\sin(x)}{x} e^{-\alpha x} \right) dx = \int_{0}^{\infty} -\sin(x) e^{-\alpha x} , dx$$
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3. Evaluating the Transformed Integral
- Using standard integration by parts twice (or the Euler identity definition of sine), we find:
- $$\int_{0}^{\infty} e^{-\alpha x} \sin(x) , dx = \frac{1}{1 + \alpha^2}$$
- Thus, the differential equation governing our parameter space is:
- $$\frac{dI}{d\alpha} = -\frac{1}{1 + \alpha^2}$$
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4. Integration and Constant Resolution
- Integrating both sides with respect to $\alpha$:
- $$I(\alpha) = -\arctan(\alpha) + C$$
- Using our asymptotic boundary condition $\lim_{\alpha \to \infty} I(\alpha) = 0$:
- $$0 = -\frac{\pi}{2} + C \implies C = \frac{\pi}{2}$$
- Therefore, the general solution for the parametric space is $I(\alpha) = \frac{\pi}{2} - \arctan(\alpha)$.
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5. Final Evaluation
- Evaluating at our target value $\alpha = 1$:
- $$I(1) = \frac{\pi}{2} - \arctan(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$