Proof of a Series for Pi

By Brad Rodgers (P1930) on Sunday, April 22, 2001 - 10:22 pm :

How would one prove

\int_{0}^{1} \frac{x^{2 n}}{1 + x^{2}} dx = \frac{π}{4} - \sum_{k = 1}^{n} \frac{(- 1)^{k + 1}}{2 k - 1}

Thanks,

Brad

By Kerwin Hui (Kwkh2) on Monday, April 23, 2001 - 03:17 pm :

Brad,

I think you missed out a minus sign somewhere. If you let $I_{n}$ be $\int_{0}^{1} x^{2 n} / (1 + x^{2}) dx$ , then

$I_{0} = π / 4$

For $n > 0$ , we have

$\begin{matrix} I_{n} & = & \int_{0}^{1} x^{2 n} / (1 + x^{2}) dx \\ = & \int_{0}^{1} x^{2 (n - 1)} (x^{2} / (1 + x^{2})) dx \\ = & \int_{0}^{1} x^{2 (n - 1)} (1 - 1 / (1 + x^{2})) dx \\ = & \int_{0}^{1} x^{2 (n - 1)} dx + \int_{0}^{1} x^{2 (n - 1)} / (1 + x^{2}) dx \\ = & 1 / (2 n - 1) - I_{n - 1} \end{matrix}$

Hence (inductively)

LHS $= I_{n} = 1 / (2 n - 1) - 1 / (2 n - 3) + \dots + (- 1)^{n} + (- 1)^{n + 1} π / 4$ .

which is not what you have written on the RHS.

Kerwin

By Brad Rodgers (P1930) on Tuesday, April 24, 2001 - 03:56 am :

I think I forgot to put an absolute value in the RHS.

Thanks,

Brad