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Question Number 136132 by frc2crc last updated on 19/Mar/21

$$Findaseriesfor\frac{x^2}{\tanh \left(x\pi \right)\tan \left(x\pi \right)}$$

Answered by Dwaipayan Shikari last updated on 19/Mar/21

$$sinh\left(\pi x\right)=\pi x\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{x^2}{n^2})$$$$\Rightarrow \frac{1}{tanh\left(\pi x\right)}=\frac{d}{dx}(log\left(\pi x\right)+\underset{n=1}{\sum ^{\mathrm{\infty }}}log(1+\frac{x^2}{n^2}))$$$$\Rightarrow \frac{1}{tanh\left(\pi x\right)}=\frac{1}{x}+2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x}{n^2+x^2}$$$$Similarly\frac{1}{tan\left(\pi x\right)}=\frac{1}{x}-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x}{n^2-x^2}$$$$\frac{x^2}{tanh\left(\pi x\right)tan\left(\pi x\right)}=1+2(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^2}{n^2+x^2}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^2}{n^2-x^2})-4x^4\left(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n^2+x^2)}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n^2-x^2)}\right)$$$$$$

Commented by frc2crc last updated on 19/Mar/21

$$iwouldhaveasumlike$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{a}_n{\left(x\pi \right)}^{2n-1}$$

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