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由Fubini定理\begin{align*}&\frac{2}{π}\int_0^∞\int_0^{π/2}\cos(x\cos θ)e^{-a x}d θd x\\&=\frac{2}{π}\int_0^{π/2}\int_0^∞\cos(x\cos θ)e^{-a x}d xd θ&\because\small\int_0^∞\cos(x\cos θ)e^{-a x}d x=ℒ\{\cos(x\cos θ)\}(a)=\frac{a}{a^2+\cos^2θ}\\&=\frac{2}{π}\int_0^{π/2}\frac{a}{a^2+\cos^2θ}d θ\\&=\frac{2a}{π}\int_0^{π/2}\frac{\sec^2θ}{a^2\sec^2θ+1}d θ&x=\tanθ\\&=\frac{2a}{π}\int_0^∞\frac1{a^2(x^2+1)+1}dx\\&=\frac{2a}{π(1+a^2)}\int_0^∞\frac1{{a^2\over a^2+1}x^2+1}dx\\&=\frac1{\sqrt{1+a^2}}
\end{align*}又见math.stackexchange.com/questions/1217543 |
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