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∫0∞e−2xsinxxdx=πa−arctanb\int_0^\infty{\dfrac{e^{-2x}\sin x}{x}}dx=\frac{\pi}a-\arctan b∫0∞xe−2xsinxdx=aπ−arctanb
The integral above holds true for the integers aaa and bbb. Find a+ba+ba+b.
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