Richard Feynman, an American physicist known for his breakthroughs in quantum mechanics, arguably developed the most elegant method to solve seemingly impossible integrals such as \( \int_{-\infty }^{\infty }\frac{\sin x}{x}dx \). Many may expect to see a complex formula or a long procedural method to solve such a problem, however, Feynman’s technique proves to be simple yet effective.
He ideated that if we introduce a parameter like a, and make the solution a function of that parameter, we can much more easily solve for the family of integrals, akin to a traditional antiderivative. Though this seems like it’s overcomplicating the problem, it truly is not. Undoubtedly, the most difficult part of the Feynman technique is identifying the parameter and its placement. We must ensure that there exists a value, or limit, of the parameter that returns the original integral. Additionally, we must ensure that there is another value, or limit, of the parameter that provides a solvable integral. Finally, taking the derivative of the entire integrand should give us an integral that is solvable with traditional methods. In the case of \( \int_{-\infty }^{\infty }\frac{\sin x}{x}dx = 2 \int_{0 }^{\infty }\frac{\sin x}{x}dx\), we cannot solve it using any traditional method due to the x being in the denominator. Instead, we can choose the parameter a within the function \( e^{-ax} \) to get the function of integration as \( I(a)=\int_{0 }^{\infty }\frac{e^{-ax}\sin x}{x}dx \). For \( a=0 \), we get back the original integral so our final solution must satisfy \( 2I(0) \). For \( \lim_{a \to \infty} I(a) \), we get \( \int_{0 }^{\infty }\frac{(0)\sin x}{x}dx=0 \) since the entire function collapses to 0 as a tends to infinity.
Then, we take the derivative of the integral with respect to the parameter. So, \( I'(a)=\frac{\mathrm{d} }{\mathrm{d} a}\int_{0}^{\infty}\frac{e^{-ax}\sin x}{x}\mathrm{dx}=\int_{0}^{\infty}\frac{\partial }{\partial x}\frac{e^{-ax}\sin x}{x}\mathrm{dx}=\)
\( \int_{0}^{\infty}\frac{-xe^{-ax}\sin x}{x}\mathrm{dx}=-\int_{0}^{\infty}e^{-ax}\sin x\mathrm{dx} \). Now that we have \( I{}'(a) \) as a solvable integral using integration by parts, we can compute the family of integrals that correspond with the derivative of our original integral.
\( -\int_{0}^{\infty}e^{-ax}\sin x\mathrm{dx}=-\left [ \frac{e^{-ax}(\cos x+a\sin x)}{1+a^{2}} \right ]_{0}^{\infty}=-(0-\frac{1+0}{1+a^{2}})=\frac{1}{1+a^{2}} \)Since \( I'(a)=\frac{1}{1+a^{2}} \), we must integrate \( I{}'(a) \) to get back our original integral.
\( \int I'(a)da=\int \frac{1}{1+a^{2}}da=\arctan (a)+C=I(a) \)Since we know \( \lim_{a \to \infty} I(a)=0 \), then \( \lim_{a \to \infty} \arctan (a) + C \) must also equal 0. Solving for \( C \), we get \( C=-\frac{\pi}{2} \)
Our final solution is \( 2I(0)= 2(\arctan (0)-\frac{\pi}{2})=\pi \)
So, \( \int_{-\infty }^{\infty }\frac{\sin x}{x}dx = \pi \) using a simple technique that Feynman introduced to the world of mathematics.
