How do I show that: ∫0∞sin⁡(ax)sin⁡(bx)x2dx=πmina,b2

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How do I show that:  ∫0∞sin⁡(ax)sin⁡(bx)x2dx=πmina,b2

Kindlein6h · Accepted Answer

In several steps:  Trigonometric relation  (sin⁡axx)(sin⁡bxx)=1−cos⁡(a+b)x2x2−1−cos⁡(a−b)x2x2  Dirchlet integral  ∫0∞sin⁡αttdt=π2sgn(α)  Integration by parts  ∫0∞1−cos⁡(α)tt2=α∫0∞sin⁡αttdt=π2|α|  Combine  ∫0∞(sin⁡axx)(sin⁡bxx)dx=π4(|a+b|−|a−b|)  =π2min(a,b)

Ella Williams · Accepted Answer

One way is to to this by residues. Another way to integrate once by parts to get  I=∫0∞bsin⁡axcos⁡bx+acos⁡axsin⁡bxxdx  then to use the formula 2sin⁡αcos⁡β=sin⁡(α+β)+sin⁡(α−β) and the mentioned integral (note that your formula needs to be corrected on the left and on the right)  ∫0∞sin⁡xyxdx=π2sgn(y)  This gives  I=π4[bsign(a+b)+bsign(a−b)+asign(a+b)+asign(b−a)]=π4(|a+b|−|a−b|)  For a,b&amp;gt;0 the last expression is obviously equal to πmina,b2

star233 · Accepted Answer

Note that   ∫0∞sin⁡(ax)xsin⁡(bx)xdx=12b∫−∞∞sin⁡(ax)axsin⁡([b/a]ax)(b/a)axadx=12b∫−∞∞sin⁡(x)xsin⁡(x/μ)x/μdx, μ≡ab&amp;gt;0 (1)∫−∞∞sin⁡(x)xsin⁡(x/μ)x/μdx|μ&amp;gt;0=∫−∞∞(12∫−11eikxdk)(12∫−11e−iqx/μdq)dx=12π∫−11∫−11∫−∞∞ei(k−q/μ)xdx2πdkdq=12π∫−11∫−11δ(k−q/u)dkdq=12π∫−11[−1&amp;lt;qμ&amp;lt;1]dq=π∫01[q&amp;lt;μ]dq=π([μ&amp;lt;1]∫0μdq+[μ&amp;gt;1]∫01dq)=[a&amp;lt;b]πab+[a&amp;gt;b]π (2)With (1) and (2):∫0∞sin⁡(ax)xsin⁡(bx)xdx=12π([a&amp;lt;b]a+[a&amp;gt;b]b)=12πmin[a,b]

How do I show that: \int_0^\infty\frac{\sin(ax)\sin(bx)}{x^2}dx=\pi min(a,b)/2

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