 | ■No20194に返信(yesterdayさんの記事)
> f(x)=1+∫[-π/2→π/2](x+sint)f(t)dtを満たす関数f(x)を求めよ。
f(x)= 1+∫[-π/2→π/2] (x+sint)f(t) dt = 1+x∫[-π/2→π/2] f(t) dt+∫[-π/2→π/2] sint・f(t) dt
∫[-π/2→π/2] f(t) dt = a …①
∫[-π/2→π/2] sint・f(t) dt = b …②
とおくと、f(x) = 1+ax+b = ax+(1+b) で
①：a = ∫[-π/2→π/2] {at+(1+b)} dt
②：b = ∫[-π/2→π/2] {at+(1+b)}sint dt 右辺は計算
a,b の連立方程式を解く。
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