← 2017 Paper 2
UPSC 2017 Maths Optional Paper 2 Q2b — Step-by-Step Solution 15 marks · Section A
Contour integration of real integrals using residues · Complex Analysis · asked 10× in 14 yrs · Read the full method →
Question
Using contour integral method, prove that ∫ 0 ∞ x sin m x a 2 + x 2 d x = π 2 e − m a \displaystyle\int_0^\infty \frac{x\sin mx}{a^2+x^2}\,dx=\frac{\pi}{2}e^{-ma} ∫ 0 ∞ a 2 + x 2 x sin m x d x = 2 π e − ma
Throughout take m > 0 m>0 m > 0 a > 0 a>0 a > 0
Technique
Upper-half-plane semicircle; replace sin m x \sin mx sin m x Im e i m x \operatorname{Im}e^{imx} Im e im x m > 0 m>0 m > 0 z = i a z=ia z = ia
Solution
Step 1 — Symmetrize and pass to a complex exponential
The integrand x sin m x a 2 + x 2 \dfrac{x\sin mx}{a^2+x^2} a 2 + x 2 x sin m x even (odd × \times ×
I = ∫ 0 ∞ x sin m x a 2 + x 2 d x = 1 2 ∫ − ∞ ∞ x sin m x a 2 + x 2 d x . I=\int_0^\infty\frac{x\sin mx}{a^2+x^2}\,dx=\frac12\int_{-\infty}^{\infty}\frac{x\sin mx}{a^2+x^2}\,dx. I = ∫ 0 ∞ a 2 + x 2 x sin m x d x = 2 1 ∫ − ∞ ∞ a 2 + x 2 x sin m x d x . Write sin m x = Im e i m x \sin mx=\operatorname{Im}e^{imx} sin m x = Im e im x
J = ∫ − ∞ ∞ x e i m x a 2 + x 2 d x , so that ∫ − ∞ ∞ x sin m x a 2 + x 2 d x = Im J , I = 1 2 Im J . J=\int_{-\infty}^{\infty}\frac{x\,e^{imx}}{a^2+x^2}\,dx,\qquad \text{so that}\quad \int_{-\infty}^\infty\frac{x\sin mx}{a^2+x^2}\,dx=\operatorname{Im}J,\quad I=\tfrac12\operatorname{Im}J. J = ∫ − ∞ ∞ a 2 + x 2 x e im x d x , so that ∫ − ∞ ∞ a 2 + x 2 x sin m x d x = Im J , I = 2 1 Im J . We choose e i m x e^{imx} e im x sin \sin sin e i m z e^{imz} e im z upper half-plane when m > 0 m>0 m > 0
Step 2 — Contour and poles
Let ϕ ( z ) = z e i m z a 2 + z 2 = z e i m z ( z − i a ) ( z + i a ) \phi(z)=\dfrac{z\,e^{imz}}{a^2+z^2}=\dfrac{z\,e^{imz}}{(z-ia)(z+ia)} ϕ ( z ) = a 2 + z 2 z e im z = ( z − ia ) ( z + ia ) z e im z C R = [ − R , R ] ∪ Γ R C_R=[-R,R]\cup\Gamma_R C R = [ − R , R ] ∪ Γ R Γ R \Gamma_R Γ R ∣ z ∣ = R |z|=R ∣ z ∣ = R Im z ≥ 0 \operatorname{Im}z\ge0 Im z ≥ 0 z = ± i a z=\pm ia z = ± ia z = i a z=ia z = ia C R C_R C R R > a R>a R > a
Step 3 — Jordan’s lemma kills the arc
Write ϕ ( z ) = z a 2 + z 2 e i m z \phi(z)=\dfrac{z}{a^2+z^2}\,e^{imz} ϕ ( z ) = a 2 + z 2 z e im z Γ R \Gamma_R Γ R
∣ z a 2 + z 2 ∣ ≤ R R 2 − a 2 → R → ∞ 0 , \left|\frac{z}{a^2+z^2}\right|\le\frac{R}{R^2-a^2}\xrightarrow[R\to\infty]{}0, a 2 + z 2 z ≤ R 2 − a 2 R R → ∞ 0 , so g ( z ) : = z a 2 + z 2 → 0 g(z):=\dfrac{z}{a^2+z^2}\to0 g ( z ) := a 2 + z 2 z → 0 Γ R \Gamma_R Γ R R → ∞ R\to\infty R → ∞ Jordan’s lemma (valid since m > 0 m>0 m > 0 g → 0 g\to0 g → 0
∫ Γ R g ( z ) e i m z d z → R → ∞ 0. \int_{\Gamma_R}g(z)e^{imz}\,dz\xrightarrow[R\to\infty]{}0. ∫ Γ R g ( z ) e im z d z R → ∞ 0. Step 4 — Residue at z = i a z=ia z = ia
For the simple pole z = i a z=ia z = ia
Res z = i a ϕ = z e i m z z + i a ∣ z = i a = i a e i m ( i a ) i a + i a = i a e − m a 2 i a = e − m a 2 . \operatorname*{Res}_{z=ia}\phi=\frac{z\,e^{imz}}{z+ia}\Bigg|_{z=ia}=\frac{ia\,e^{im(ia)}}{ia+ia}=\frac{ia\,e^{-ma}}{2ia}=\frac{e^{-ma}}{2}. z = ia Res ϕ = z + ia z e im z z = ia = ia + ia ia e im ( ia ) = 2 ia ia e − ma = 2 e − ma . (Used e i m ( i a ) = e − m a e^{im(ia)}=e^{-ma} e im ( ia ) = e − ma
Step 5 — Residue theorem and limit
For R > a R>a R > a
∮ C R ϕ d z = 2 π i Res z = i a ϕ = 2 π i ⋅ e − m a 2 = π i e − m a . \oint_{C_R}\phi\,dz=2\pi i\operatorname*{Res}_{z=ia}\phi=2\pi i\cdot\frac{e^{-ma}}{2}=\pi i\,e^{-ma}. ∮ C R ϕ d z = 2 π i z = ia Res ϕ = 2 π i ⋅ 2 e − ma = π i e − ma . Letting R → ∞ R\to\infty R → ∞
J = ∫ − ∞ ∞ x e i m x a 2 + x 2 d x = π i e − m a . J=\int_{-\infty}^{\infty}\frac{x\,e^{imx}}{a^2+x^2}\,dx=\pi i\,e^{-ma}. J = ∫ − ∞ ∞ a 2 + x 2 x e im x d x = π i e − ma . Take the imaginary part:
∫ − ∞ ∞ x sin m x a 2 + x 2 d x = Im ( π i e − m a ) = π e − m a . \int_{-\infty}^{\infty}\frac{x\sin mx}{a^2+x^2}\,dx=\operatorname{Im}\big(\pi i\,e^{-ma}\big)=\pi e^{-ma}. ∫ − ∞ ∞ a 2 + x 2 x sin m x d x = Im ( π i e − ma ) = π e − ma . (As a bonus, Re J = ∫ x cos m x a 2 + x 2 d x = 0 \operatorname{Re}J=\int\frac{x\cos mx}{a^2+x^2}dx=0 Re J = ∫ a 2 + x 2 x c o s m x d x = 0
I = 1 2 ⋅ π e − m a = π 2 e − m a . I=\frac12\cdot\pi e^{-ma}=\boxed{\;\frac{\pi}{2}\,e^{-ma}.\;} I = 2 1 ⋅ π e − ma = 2 π e − ma . ■ \blacksquare ■
Verification
Numerical (SciPy quad, oscillatory weight='sin'):
m = 1 , a = 1 m=1,a=1 m = 1 , a = 1 0.5755 0.5755 0.5755 π 2 e − 1 = 0.5779 \frac{\pi}{2}e^{-1}=0.5779 2 π e − 1 = 0.5779 m = 2 , a = 1 m=2,a=1 m = 2 , a = 1 0.2139 0.2139 0.2139 π 2 e − 2 = 0.2126 \frac{\pi}{2}e^{-2}=0.2126 2 π e − 2 = 0.2126 m = 1 , a = 2 m=1,a=2 m = 1 , a = 2 0.2102 0.2102 0.2102 π 2 e − 2 = 0.2126 \frac{\pi}{2}e^{-2}=0.2126 2 π e − 2 = 0.2126 m = 3 , a = 0.5 m=3,a=0.5 m = 3 , a = 0.5 0.3522 0.3522 0.3522 π 2 e − 1.5 = 0.3505 \frac{\pi}{2}e^{-1.5}=0.3505 2 π e − 1.5 = 0.3505
Residue cross-check: Res i a ϕ = e − m a / 2 \operatorname{Res}_{ia}\phi=e^{-ma}/2 Res ia ϕ = e − ma /2 2 π i ⋅ e − m a / 2 = π i e − m a 2\pi i\cdot e^{-ma}/2=\pi i e^{-ma} 2 π i ⋅ e − ma /2 = π i e − ma Im = π e − m a \operatorname{Im}=\pi e^{-ma} Im = π e − ma = π 2 e − m a =\frac{\pi}{2}e^{-ma} = 2 π e − ma