Sum of Reciprocals of Squares of Odd Integers/Proof 3

Theorem

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)

\(=\)

\(\ds 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots\)

\(\ds \)

\(=\)

\(\ds \dfrac {\pi^2} 8\)

Proof

This "proof" leads to a dead end. Its value lies in showing that this approach does not work.

\(\ds \)

\(\)

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)

\(\ds \)

\(=\)

\(\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y\)

Sum of Reciprocals of Squares of Odd Integers as Double Integral

\(\ds \)

\(=\)

\(\ds \int_0^1 \int_0^1 \frac 1 {\paren {1 + x y} \paren {1 - x y} } \rd x \rd y\)

\(\ds \)

\(=\)

\(\ds \int_0^1 \int_0^1 \frac 1 {2 \paren {1 + x y} } \rd x \rd y - \int_0^1 \int_0^1 \frac 1 {2 \paren {1 - x y} } \rd x \rd y\)

Primitive of Function of Constant Multiple, partial fraction expansion

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int_0^1 \intlimits {\frac {\map \ln {1 + x y} } x} 0 1 \rd x - \frac 1 2 \int_0^1 \intlimits {\frac {\map \ln {1 - x y} } x} 0 1 \rd x\)

Primitive of Reciprocal

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int_0^1 \paren {\frac {\map \ln {1 + x} } x - \frac {\ln 1} x} \rd x - \frac 1 2 \int_0^1 \paren {\frac {\map \ln {1 - x} } x - \frac {\ln 1} x} \rd x\)

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int_0^1 \frac 1 x \map \ln {\frac {1 + x} {1 - x} } \rd x\)

Sum of Logarithms

\(\ds \)

\(=\)

\(\ds \int_0^1 \frac {\artanh x} x \rd x\)

Definition of Real Area Hyperbolic Tangent

\(\ds \)

\(=\)

\(\ds \int_0^1 \frac 1 x \paren {x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots} \rd x\)

Power Series Expansion for Real Area Hyperbolic Tangent

\(\ds \)

\(=\)

\(\ds \int_0^1 \paren {1 + \frac {x^2} 3 + \frac {x^4} 5 + \frac {x^6} 7 + \cdots} \rd x\)

\(\ds \)

\(=\)

\(\ds \intlimits {x + \frac {x^3} 9 + \frac {x^5} {25} + \frac {x^7} {49} + \cdots} 0 1\)

\(\ds \)

\(=\)

\(\ds 1 + \frac 1 9 + \frac 1 {25} + \frac 1 {49} + \cdots\)

and a dead end is reached