calculation , residue theorem
Use the residue theorem to prove that
∫ 0 ∞ e cos x sin ( sin x ) d x x = π 2 ( e − 1 ) . \int_0^\infty e^{\cos{x}} \sin\p{\sin{x}} \,\frac{\diff{x}}{x}
= \frac{\pi}{2}\p{e - 1}. ∫ 0 ∞ e c o s x sin ( sin x ) x d x = 2 π ( e − 1 ) .
Let f ( z ) = e e i z z f\p{z} = \frac{e^{e^{iz}}}{z} f ( z ) = z e e i z x ∈ R x \in \R x ∈ R
f ( x ) = e cos x + i sin x x = e cos x ( cos ( sin x ) + i sin ( sin x ) ) x , f\p{x}
= \frac{e^{\cos{x} + i\sin{x}}}{x}
= \frac{e^{\cos{x}}\p{\cos\p{\sin{x}} + i\sin\p{\sin{x}}}}{x}, f ( x ) = x e c o s x + i s i n x = x e c o s x ( cos ( sin x ) + i sin ( sin x ) ) , so the imaginary part of f ( x ) f\p{x} f ( x ) f f f f f f
Res ( f ; 0 ) = lim z → 0 z f ( z ) = lim z → 0 e e i z = e . \Res{f}{0}
= \lim_{z\to0} zf\p{z}
= \lim_{z\to0} e^{e^{iz}}
= e. Res ( f ; 0 ) = z → 0 lim z f ( z ) = z → 0 lim e e i z = e . Observe also that if z = x + i y z = x + iy z = x + i y e i z = e − y ( cos x + i sin x ) e^{iz} = e^{-y}\p{\cos{x} + i\sin{x}} e i z = e − y ( cos x + i sin x )
∣ e e i z ∣ = e Re e i z = e e − y cos x . \abs{e^{e^{iz}}}
= e^{\Re{e^{iz}}}
= e^{e^{-y}\cos{x}}. ∣ ∣ e e i z ∣ ∣ = e Re e i z = e e − y c o s x . For 0 < ε < R 0 < \epsilon < R 0 < ε < R γ ε , R \gamma_{\epsilon,R} γ ε , R [ − R , − ε ] \br{-R, -\epsilon} [ − R , − ε ] [ ε , R ] \br{\epsilon, R} [ ε , R ] { R e i θ ∣ 0 ≤ θ ≤ π } \set{Re^{i\theta} \mid 0 \leq \theta \leq \pi} { R e i θ ∣ 0 ≤ θ ≤ π } { ε e i θ ∣ π ≤ θ ≤ 2 π } \set{\epsilon e^{i\theta} \mid \pi \leq \theta \leq 2\pi} { ε e i θ ∣ π ≤ θ ≤ 2 π } 0 0 0 γ ε , R \gamma_{\epsilon,R} γ ε , R C R C_R C R
∣ f ( z ) ∣ = e e − R sin θ cos ( cos θ ) R ≤ e R ≤ e , \begin{aligned}
\abs{f\p{z}}
= \frac{e^{e^{-R\sin\theta}\cos\p{\cos\theta}}}{R}
\leq \frac{e}{R}
\leq e,
\end{aligned} ∣ f ( z ) ∣ = R e e − R s i n θ c o s ( c o s θ ) ≤ R e ≤ e , since sin θ ≥ 0 \sin\theta \geq 0 sin θ ≥ 0
lim R → ∞ ∫ C R f ( z ) d z = i ∫ 0 π lim R → ∞ e e − i R e i θ d θ = i ∫ 0 π e 0 d θ = i π . \lim_{R\to\infty} \int_{C_R} f\p{z} \,\diff{z}
= i\int_0^\pi \lim_{R\to\infty} e^{e^{-iRe^{i\theta}}} \,\diff\theta
= i\int_0^\pi e^0 \,\diff\theta
= i\pi. R → ∞ lim ∫ C R f ( z ) d z = i ∫ 0 π R → ∞ lim e e − i R e i θ d θ = i ∫ 0 π e 0 d θ = iπ . On the lower arc oriented clockwise, observe that because f f f z = 0 z = 0 z = 0 f ( z ) − e z f\p{z} - \frac{e}{z} f ( z ) − z e 0 0 0
∫ C ε f ( z ) d z = ∫ C ε ( f ( z ) − e z ) + e z d z = ∫ C ε ( f ( z ) − e z ) d z + i π e → ε → 0 i π e , \begin{aligned}
\int_{C_\epsilon} f\p{z} \,\diff{z}
&= \int_{C_\epsilon} \p{f\p{z} - \frac{e}{z}} + \frac{e}{z} \,\diff{z} \\
&= \int_{C_\epsilon} \p{f\p{z} - \frac{e}{z}} \,\diff{z} + i\pi e
\xrightarrow{\epsilon\to0} i\pi e,
\end{aligned} ∫ C ε f ( z ) d z = ∫ C ε ( f ( z ) − z e ) + z e d z = ∫ C ε ( f ( z ) − z e ) d z + iπ e ε → 0 iπ e , since f ( z ) − e z f\p{z} - \frac{e}{z} f ( z ) − z e 0 0 0 ε → 0 \epsilon \to 0 ε → 0 R → ∞ R \to \infty R → ∞
i π + i π e + ∫ − ∞ ∞ f ( x ) d x = 2 π i e ⟹ ∫ 0 ∞ e cos x sin ( sin x ) d x x = π 2 ( e − 1 ) , i\pi + i\pi e + \int_{-\infty}^\infty f\p{x} \,\diff{x} = 2\pi i e
\implies \int_0^\infty e^{\cos{x}}\sin\p{\sin{x}} \,\frac{\diff{x}}{x} = \frac{\pi}{2}\p{e - 1}, iπ + iπ e + ∫ − ∞ ∞ f ( x ) d x = 2 πi e ⟹ ∫ 0 ∞ e c o s x sin ( sin x ) x d x = 2 π ( e − 1 ) , since the integrand is even, and by taking imaginary parts.
Solution.