A Sinful Double Improper Integral

Calculus Level 4

Round the exact value of ∫ 0 ∞ ∫ 0 ∞ sin ⁡ ( x ) sin ⁡ ( x + y ) x ( x + y ) d y d x \int_0^{\infty} \int_0^{\infty} \frac{\sin(x)\sin(x+y)}{x(x+y)}\,\mathrm dy\,\mathrm dx to the nearest hundredths place.

Hint: Contour Integration and Complex-Analytic methods were my preferred approaches to solving this problem, although there is another short-cut.

1.20 1.21 1.22 1.23 1.25 1.24

Sanjoy Kundu by Sanjoy Kundu , 22, USA

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1 solution

taking x + y = t x+y = t

Integral becomes ∫ 0 ∞ s i n x x ( ∫ x ∞ s i n t t d t ) d x \mathrm{\int_0^∞ \frac{sinx}{x}( \int_x^∞ \frac{sin t}{t} dt) dx } ∫ 0 ∞ s i n x x ( ∫ 0 ∞ s i n t t − ∫ 0 x s i n t t d t ) d x \mathrm{\int_0^∞ \frac{sinx}{x} (\int_0^∞ \frac{sin t}{t} -\int_0^x \frac{sin t}{t}dt) dx} = π 2 × π 2 − ∫ 0 ∞ S i ( x ) s i n x x = \mathrm{\frac {π}{ 2}×\frac{π}{2} - \int_0^∞ Si(x) \frac{sinx}{x}} = π 2 4 − ∫ 0 ∞ S i ( x ) S i ′ ( x ) d x = \mathrm{\frac{π^2}{4} -\int_0^∞ Si(x) Si'(x)dx} = π 2 4 − 1 2 lim ⁡ x → ∞ [ S i 2 ( x ) ] = π 2 8 = 1.233 = \mathrm{\frac{π^2}{4} -\frac{1}{2} \lim_{x\rightarrow{∞}}[Si^2 (x)] }= \color{#20A900}\boxed{\frac{π^2}{8} =1.233}

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