>>198
>>145なんて只の有理関数の積分で面倒くさいだけだろ
∫f(t)g(x−t)dt＝∫(1/(t^2＋a^2))(1/((x−t)^2＋b^2))dt
＝(1/((x^2＋(a−b)^2)(x^2＋(a＋b)^2)))∫(2x((x−t)/((x−t)^2＋b^2)＋t/(t^2＋a^2))＋(x^2＋a^2−b^2)/((x−t)^2＋b^2)＋(x^2−a^2＋b^2)/(t^2＋a^2))dt
＝(1/((x^2＋(a−b)^2)(x^2＋(a＋b)^2)))(x ln((t^2＋a^2)/((x−t)^2＋b^2))＋(1/a)(x^2−a^2＋b^2)arctan(t/a)−(1/b)(x^2＋a^2−b^2)arctan((x−t)/b))
留数の方が良かったかな？