# F(x)=a+b(arctan(x/2)) ``Fˇ(x)=f(x) intergal from -infinity to infinity of f(x)=1 How to find a and b ? Help please

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### 2 Answers

You need to use the fact that `int f(x) dx = F(x)` , hence, since the problem provides the information that `int_(-oo)^(oo) f(x) dx = 1` , hence `F(x)|_(-oo)^(oo) = 1` , such that:

`F(x)|_(-oo)^(oo) = F(oo) - F(-oo)`

`F(x)|_(-oo)^(oo) = a + b arctan (oo/2) - a - b arctan (-oo/2)`

You need to use the following identity, such that:

`arctan(- alpha) = -arctan alpha`

`F(x)|_(-oo)^(oo) = b(arctan (oo/2) - (-arctan (oo/2)))`

`F(x)|_(-oo)^(oo) = b*2arctan (oo)`

Since `arctan oo = pi/2` yields:

`F(x)|_(-oo)^(oo) = b*2(pi/2) = b*pi`

Since `int_(-oo)^(oo) f(x) dx = 1` , hence `b*pi = 1 => b = 1/pi.`

**Hence, evaluating a and b, under the given conditions, yields `a in R` and **` b = 1/pi.`

here derivative of F(x) is f(x)