Proving $\sum_{k=0} \frac{(-1)^k}{1+2k} = \frac{\pi}{4} $ taking $4 \int_0^1 \sqrt{1 - x^2} \; \text{d}x$ as the definition of $\pi$.

Question

I'm trying to prove the Madhava-Leibniz formula for $\pi$ using one quarter of the area of the unit circle as the definition of $\frac{\pi}{4}$.

$$ \int_{0}^{1} \sqrt{1-xx}\; \text{d}x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} $$

I have a proof of this identity, but it uses a lot of trigonometric substitutions and makes direct use of $\pi$ as a constant.

This feels a bit circular since the left hand side is supposed to be the definition of $\frac{\pi}{4}$.

Is there a way to prove this identity directly, instead of proving that both sides are equal to $\frac{\pi}{4}$?

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All integrals are over the interval $[0,1]$ in the variable $x$. Let $\theta$ refer to $\frac{\pi}{4} x$. $\theta$ is a constant multiple of $x$, I'm using it so I don't need to change the bound of integration.

$$ \sum_{k=0} \frac{(-1)^k}{1+2k} $$

Using Lemma 101.

$$ \sum_{k=0} \int_{0}^{1} (-1)^k x^{2k} \text{d}x $$ $$ \sum_{k=0} \int_{0}^{1} (-xx)^k \text{d}x $$

Exchanging the sum and integral.

$$ \int_{0}^{1} \sum_{k=0} (-xx)^k \text{d}x $$

Applying Lemma 102

$$ \int_{0}^{1} \frac{1}{1+xx} \text{d}x $$

Applying Lemma 103, with $g$ equal to $\tan(\theta)$ and $g'$ equal to $\frac{\pi}{4} \sec^2(\theta)$.

$$ \int_{0}^{1} \frac{\pi}{4} \cdot \sec^2(\theta) \cdot \frac{1}{1+\tan^2(\theta)} \text{d}x $$

$$ \frac{\pi}{4} \int_{0}^{1} \sec^2(\theta) \cdot \frac{1}{\sec^2(\theta)} \text{d}x $$

$$ \frac{\pi}{4} \int_{0}^{1} 1\; \text{d}x $$

$$ \frac{\pi}{4} $$

And for the other side.

$$ \int_0^1 \sqrt{1-xx} \; \text{d}x $$

Apply lemma 103, with $g = \sin(2\theta)$ and $g' = \frac{\pi}{2}\cos(2\theta)$.

$$ \int_0^1 \frac{\pi}{2} \cdot \cos(2 \theta) \sqrt{1-\sin^2(2 \theta)} \; \text{d}x $$

$$ \frac{\pi}{2} \int_0^1 \cos^2(2 \theta) \; \text{d}x $$

Apply lemma 104.

$$ \frac{\pi}{2} \cdot \frac{1}{2} $$

$$ \frac{\pi}{4} $$

Lemmas

Lemma 101: integral of monomials

$$ \int_0^1 x^n \text{d}x = \frac{1}{1+n} $$

Lemma 102: sum of geometric series

$$ \sum_{k=0} r^k = \frac{1}{1-r} $$

Lemma 103: Special case of change of variables for integration

Suppose $g(0)=0$ and $g(1)=1$, then the following holds.

$$ \int_0^1 f(x) \text{d}x = \int_0^1 g'(x) \cdot f'\circ g(x) \text{d}x $$

Lemma 104: Integral of $\cos^2 + \sin^2$

For all values of $x$, $\cos^2(x) + \sin^2(x) = 1$, therefore the following holds

$$ \int_0^1 \cos^2(2\theta) + \sin^2(2\theta) \text{d}x = 1 $$

However, $\cos\left(\frac{\pi}{2}-x\right) = \sin(x) $, therefore, over the interval $[0, 1]$, $\sin$ and $\cos$ take on the same values at the same offsets, but in the opposite order. Thus

$$ \int_0^1 \sin^2(2\theta) \text{d}x = \int_0^1 \cos^2(2\theta) \text{d}x $$

Thus

$$ \int_0^1 \cos^2(2\theta) \text{d}x = \frac{1}{2} $$

Answer 1

Substitute $x=t/\sqrt{1+t^2}$; then $dx=dt/(1+t^2)^{3/2}$ and $$\int_0^1\sqrt{1-x^2}\,dx=\int_0^\infty\frac{dt}{(1+t^2)^2}=\int_0^1\frac{dt}{(1+t^2)^2}+\int_0^1\frac{dt/t^2}{(1+1/t^2)^2}=\int_0^1\frac{dt}{1+t^2}.$$ Now you know what to do next.

Answer 2

The right hand side of your formula is the Maclaurin series expansion of $\arctan 1$.

Answer 3

Too long for coments.

Considering $$S_{m,n}=\sum_{k=m}^{n} \frac{(-1)^k}{2k+1}$$ $$S_{m,n}=\frac{1}{2} \left((-1)^m \,\Phi \left(-1,1,m+\frac{1}{2}\right)+(-1)^n\, \Phi\left(-1,1,n+\frac{3}{2}\right)\right)$$ where appears the Lerch transcendent function.

$$S_{0,n}=\frac{1}{4} \left(2 (-1)^n \Phi \left(-1,1,n+\frac{3}{2}\right)+\psi \left(\frac{3}{4}\right)-\psi \left(\frac{1}{4}\right)\right)$$ $$\psi \left(\frac{3}{4}\right)-\psi \left(\frac{1}{4}\right)=\pi$$ $$S_{0,n}=\frac{1}{2} (-1)^n \Phi \left(-1,1,n+\frac{3}{2}\right)+\frac{\pi }{4}$$ $$S_{0,n}=\frac{1}{4} (-1)^n \left(\psi \left(\frac{2 n+5}{4} \right)-\psi \left(\frac{2 n+3}{4} \right)\right)+\frac{\pi }{4}$$ Using asymptotics $$S_{0,n}=\frac{\pi }{4}+\frac{(-1)^n }{4n} \left(1-\frac{1}{n}+\frac{3}{4 n^2}+O\left(\frac{1}{n^3}\right)\right)$$