We want to prove the following improper integral

Theorem.

lim⁡a→∞∫0asin⁡xx dx=π2. \lim_{a \rightarrow \infty} \int_0^a \frac{\sin x}{x} \, dx = \frac{\pi}{2}.

Existence of Limit

Since ∫nπ(n+1)πsin⁡x dx=(−1)n⋅2\int_{n \pi}^{(n + 1) \pi} \sin x \, dx = (-1)^{n} \cdot 2,

∫nπ(n+1)πsin⁡xx dx∈Θ((−1)nn). \int_{n \pi}^{(n + 1) \pi} \frac{\sin x}{x} \, dx \in \Theta \left( \frac{ (-1)^{n} }{n} \right).

Hence lim⁡a→∞∫0asin⁡xx dx\lim_{a \rightarrow \infty} \int_0^a \frac{\sin x}{x} \, dx exists. On the other hand, sin⁡xx\frac{\sin x}{x} is not Lebesgue integrable over [0,∞)[0, \infty).

Proving the Limit

Observation.

∫0∞e−xysin⁡x dy=sin⁡xx. \int_0^\infty e^{-xy} \sin x \, dy = \frac{\sin x}{x}.

We need two technical lemmas to finish the proof.

Lemma 1. e−xysin⁡xe^{-xy} \sin x is integrable on [0,a]×[0,∞)[0, a] \times [0, \infty).

Lemma 2.

∫0ae−xysin⁡x dx=−e−xycos⁡x∣0a−ye−xysin⁡x∣0a1+y2=1−e−aycos⁡a−ye−aysin⁡a1+y2. \begin{aligned} \int_0^a e^{-xy} \sin x \, dx &= \frac{ - e^{-xy} \cos x \mid_0^a - y e^{-xy} \sin x \mid_0^a }{1 + y^2} \\ &= \frac{ 1 - e^{- ay} \cos a - y e^{-ay} \sin a }{1 + y^2}. \end{aligned}

We defer the proof of the lemmas to the end of the article. Denote F\mathcal{F} the collection of Lebesgue measurable sets on R\mathbb{R}, and λ\lambda the Lebesgue measure defined on F\mathcal{F}. Let μ≐λ×λ\mu \doteq \lambda \times \lambda be the product measure on (R×R,F⊗F)(\mathbb{R} \times \mathbb{R}, \mathcal{F} \otimes \mathcal{F}). Now, via Fubini's theorem,

∫0asin⁡xxdx=∫0a∫0∞e−xysin⁡x dydx=∫[0,a]×[0,∞)e−xysin⁡x dμ=∫0∞∫0ae−xysin⁡x dxdy=∫0∞1−e−aycos⁡a−ye−aysin⁡a1+y2 dy=arctan⁡y∣0∞−∫0∞e−aycos⁡a+ye−aysin⁡a1+y2 dy=π2−∫0∞e−aycos⁡a+ye−aysin⁡a1+y2 dy. \begin{aligned} \int_0^a \frac{\sin x}{x} dx &= \int_0^a \int_0^\infty e^{-xy} \sin x \, d y dx \\ &=\int_{[0, a] \times [0, \infty)} e^{-xy} \sin x \, d \mu \\ &= \int_0^\infty \int_0^a e^{-xy} \sin x \, dx d y \\ &=\int_0^\infty \frac{ 1 - e^{- ay} \cos a - y e^{-ay} \sin a }{1 + y^2} \, d y \\ &= \arctan y \mid_0^\infty - \int_0^\infty \frac{ e^{- ay} \cos a + y e^{-ay} \sin a }{1 + y^2} \, d y \\ &= \frac{\pi}{2} - \int_0^\infty \frac{ e^{- ay} \cos a + y e^{-ay} \sin a }{1 + y^2} \, d y. \end{aligned}

The right hand side converges to 00 as aa approaches infinity:

lim⁡a→∞∫0∞∣e−aycos⁡a+ye−aysin⁡a1+y2∣ dy≤lim⁡a→∞∫0∞(1+y)e−ay dy=lim⁡a→∞−1a((1+y)e−ay+1ae−ay)∣0∞=0. \begin{aligned} \lim_{a \rightarrow \infty} \int_0^\infty \left| \frac{ e^{- ay} \cos a + y e^{-ay} \sin a }{1 + y^2} \right| \, d y &\le \lim_{a \rightarrow \infty} \int_0^\infty (1 + y) e^{-ay} \, d y \\ &= \lim_{a \rightarrow \infty} -\frac{1}{a} \left( (1 + y) e^{-ay} + \frac{1}{a} e^{-ay} \right) \Bigg|_0^\infty = 0. \end{aligned}

Proof of Lemma 1. Via Tonelli's theorem,

∫[0,a]×[0,∞)∣e−xysin⁡x∣ dμ=∫0a∫0∞∣e−xysin⁡x∣ dydx≤∫0a∫0∞e−xy dydx=∫0a1x dydx=ln⁡a≤∞. \begin{aligned} \int_{[0, a] \times [0, \infty)} | e^{-xy} \sin x | \, d \mu &= \int_0^a \int_0^\infty | e^{-xy} \sin x | \, d y dx \\ &\le \int_0^a \int_0^\infty e^{-xy} \, d y dx \\ &= \int_0^a \frac{1}{x} \, d y dx \\ &= \ln a \\ &\le \infty. \end{aligned}

□\square

Proof of Lemma 2.

∫0ae−xysin⁡x dx=−e−xycos⁡x∣0a−y∫0ae−xycos⁡x dx=−e−xycos⁡x∣0a−y(e−xysin⁡x∣0a+y∫0ae−xysin⁡x dx). \begin{aligned} \int_0^a e^{-xy} \sin x \, dx &= - e^{-xy} \cos x \mid_0^a - y \int_0^a e^{-xy} \cos x \, dx \\ &= - e^{-xy} \cos x \mid_0^a - y \left( e^{-xy} \sin x \mid_0^a + y \int_0^a e^{-xy} \sin x \, dx \right). \end{aligned}

Another proof of Lemma 2.

∫0ae−xyeix dx=∫0ae−x(y−i) dx=−1y−i⋅e−x(y−i)∣0a=1y−i⋅(1−e−a(y−i))=y+i1+y2⋅(1−e−ay(cos⁡a+isin⁡a)). \begin{aligned} \int_0^a e^{-xy} e^{ix} \, dx &= \int_0^a e^{-x (y - i)} \, dx \\ &= \frac{-1}{y - i} \cdot e^{-x (y - i)} \mid_0^a \\ &= \frac{1}{y - i} \cdot \left( 1 - e^{- a(y - i)} \right) \\ &= \frac{y + i}{1 + y^2} \cdot \left( 1 - e^{-ay} (\cos a + i \sin a ) \right). \end{aligned}

Hence,

∫0ae−xysin⁡x dx=ℑ(∫0ae−xyeix dx)=1−e−aycos⁡a−ye−aysin⁡a1+y2. \int_0^a e^{-xy} \sin x \, dx = \Im \left( \int_0^a e^{-xy} e^{ix} \, dx \right) = \frac{ 1 - e^{- ay} \cos a - y e^{-ay} \sin a }{1 + y^2}.

□\square

Reference

[1] P. Billingsley, Probability and measure Anniversary Edition, 3rd ed. New York: Wiley, 1995.