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154

Part of 2006 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 154

Source: Kyoto University entrance exam/science 2002 Second League

9/27/2006
Find the function f(x)f(x) which is defined for [−π2, π2]\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right] such that f(x)+∫−π2π2sin⁡(x−y)⋅f(y) dy=x+1 (−π2≤x≤π2).f(x)+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin (x-y)\cdot f(y)\ dy=x+1\ \left(-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}\right).
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