อ้างอิง:
ข้อความเดิมของคุณ Mastermander:
78. Compute $$ \int_0^\infty \arctan\dfrac{\Theta^2}{x^2}\,dx $$
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ไม่แน่ใจว่ามีวิธีสั้นกว่านี้หรือเปล่านะครับ
Let $ u= \frac{\Theta}{x} $
And integration becomes $$ \Theta \int_0^\infty \dfrac{\arctan u^2}{u^2}\,\, du $$
$$ \begin{array}{rcl} \int_0^\infty \dfrac{\arctan u^2}{u^2}\,\, du &=& \int_0^\infty \dfrac{1}{u^2}\bigg(\int_0^u \dfrac{2v}{1+v^4}\,\, dv\bigg) \,\, du \\ &=& \int_0^\infty\!\!\! \int_0^u \dfrac{2v}{u^2(1+v^4)}\,\, dvdu \\ &=& \int_0^\infty\!\!\! \int_v^{\infty} \dfrac{2v}{u^2(1+v^4)}\,\, dudv \\&=& \int_0^\infty \dfrac{2}{1+v^4}\,\, dv \\&=&\int_0^\infty \dfrac{v^2+1}{v^4+1} - \dfrac{v^2-1}{v^4+1}\,\, dv \\&=& \int_0^\infty \dfrac{1+\frac{1}{v^2}}{v^2+\frac{1}{v^2}} - \dfrac{1-\frac{1}{v^2}}{v^2+\frac{1}{v^2}}\,\, dv \end{array} $$
First integrand : use $ w= v- \frac{1}{v}$
Second integrand : use $ z= v +\frac{1}{v}$
Finally , solution of question 78 is $ \frac{\Theta \pi}{\sqrt{2}} $