How to Integrate the Sinc Function

Step-by-Step Guide

1. 1

   Step 1: Begin with the integral to be evaluated.

   We are evaluating over the entire real line, so the limits will be positive and negative infinity.
   
   Above is a visualization of the function with both definitions
   - unnormalized (in red) and normalized (in blue).
   
   We will be evaluating the unnormalized sinc function. ∫−∞∞sin⁡xxdx{\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}{\mathrm {d} }x} We see from the graph that sin⁡xx{\displaystyle {\frac {\sin x}{x}}} is an even function, which can be confirmed by looking at the function above.
   
   Then, we can factor out a
   2. 2∫0∞sin⁡xxdx{\displaystyle 2\int _{0}^{\infty }{\frac {\sin x}{x}}{\mathrm {d} }x} The integral above with bounds of 0 to infinity is also known as the Dirichlet integral.
2. 2

   Step 2: Define a function G(t){\displaystyle G(t)}.

   The purpose of defining such a function with an argument t{\displaystyle t} is so that we can work with an integral that is easier to evaluate, whilst meeting the conditions of the sinc integral for appropriate values of t.{\displaystyle t.} In other words, putting the e−tx{\displaystyle e^{-tx}} term inside the integral is valid, since the integral converges for all t≥0,{\displaystyle t\geq 0,} while setting t=0{\displaystyle t=0} recovers the original integral.
   
   This reformulation means that we are ultimately evaluating G(0).{\displaystyle G(0).} G(t)=∫0∞sin⁡xxe−txdx{\displaystyle G(t)=\int _{0}^{\infty }{\frac {\sin x}{x}}e^{-tx}{\mathrm {d} }x} , We can move the derivative under the integration sign because the integral is being taken with respect to a different variable.
   
   While we do not justify this operation here, it is widely applicable for a great many functions.
   
   Keep in mind that t{\displaystyle t} is to be treated as a variable throughout the evaluation, not a constant. dGdt=ddt∫0∞sin⁡xxe−txdx=∫0∞∂∂te−txsin⁡xxdx=−∫0∞e−txsin⁡xdx{\displaystyle {\begin{aligned}{\frac {{\mathrm {d} }G}{{\mathrm {d} }t}}&={\frac {\mathrm {d} }{{\mathrm {d} }t}}\int _{0}^{\infty }{\frac {\sin x}{x}}e^{-tx}{\mathrm {d} }x\\&=\int _{0}^{\infty }{\frac {\partial }{\partial t}}e^{-tx}{\frac {\sin x}{x}}{\mathrm {d} }x\\&=-\int _{0}^{\infty }e^{-tx}\sin x{\mathrm {d} }x\end{aligned}}} , This is, in fact, the evaluation for the Laplace transform of sin⁡x.{\displaystyle \sin x.} The most basic way to evaluate this integral is by using integration by parts, which we work out below.
   
   See the tips for a more powerful way to integrate this.
   
   Pay attention to the signs. dGdt=e−txcos⁡x|0∞+∫0∞te−txcos⁡xdx=e−txcos⁡x|0∞+=e−txcos⁡x|0∞+te−txsin⁡x|0∞−t2dGdt{\displaystyle {\begin{aligned}{\frac {{\mathrm {d} }G}{{\mathrm {d} }t}}&=e^{-tx}\cos x{\Big |}_{0}^{\infty }+\int _{0}^{\infty }te^{-tx}\cos x{\mathrm {d} }x\\&=e^{-tx}\cos x{\Big |}_{0}^{\infty }+\left\\&=e^{-tx}\cos x{\Big |}_{0}^{\infty }+te^{-tx}\sin x{\Big |}_{0}^{\infty }-t^{2}{\frac {{\mathrm {d} }G}{{\mathrm {d} }t}}\end{aligned}}} dGdt(1+t2)=(0−1)+(0−0){\displaystyle {\frac {{\mathrm {d} }G}{{\mathrm {d} }t}}(1+t^{2})=(0-1)+(0-0)} dGdt=−11+t2{\displaystyle {\frac {{\mathrm {d} }G}{{\mathrm {d} }t}}=-{\frac {1}{1+t^{2}}}} , This recovers G(t){\displaystyle G(t)} under a different variable.
   
   Since the integrand is the differential of a well-known function, this evaluation is trivial. −∫11+t2dt=−tan−1⁡t+C{\displaystyle
   -\int {\frac {1}{1+t^{2}}}{\mathrm {d} }t=-\tan ^{-1}t+C} Here, we recognize that G(t)=0{\displaystyle G(t)=0} as t→∞{\displaystyle t\to \infty } for both this integral and the one defined in step
   2.
   
   However, limt→∞tan−1⁡t=π2,{\displaystyle \lim _{t\to \infty }\tan ^{-1}t={\frac {\pi }{2}},} so C=π2{\displaystyle C={\frac {\pi }{2}}} as well.
   
   Therefore, G(t)=−tan−1⁡t+π2.{\displaystyle G(t)=-\tan ^{-1}t+{\frac {\pi }{2}}.} , Now that we have G(t),{\displaystyle G(t),} where t≥0,{\displaystyle t\geq 0,} we can substitute 0 for t{\displaystyle t} and find that G(0)=π2.{\displaystyle G(0)={\frac {\pi }{2}}.} G(0)=∫0∞sin⁡xxdx=π2{\displaystyle G(0)=\int _{0}^{\infty }{\frac {\sin x}{x}}{\mathrm {d} }x={\frac {\pi }{2}}} Finally, we recall that to integrate over all the reals, we simply multiply by 2, as sinc⁡x{\displaystyle \operatorname {sinc} x} is an even function. ∫−∞∞sin⁡xxdx=π{\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}{\mathrm {d} }x=\pi } It is worth memorizing this answer, as it can pop up in multiple contexts.
3. 3

   Step 3: Differentiate under the integral.

4. 4

   Step 4: Evaluate dGdt{\displaystyle {\frac {{\mathrm {d} }G}{{\mathrm {d} }t}}}.

5. 5

   Step 5: Integrate both sides with respect to t{\displaystyle t}.

6. 6

   Step 6: Evaluate the sinc integral.

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