Question

$\displaystyle \sf\int_{0}^{ \infty } \frac{ {e}^{ - 2x} \: tanh(x)}{x} \: dx$​

Answer 1

Step-by-step explanation:

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Answer 2

$\displaystyle \tt \red{\int_{0}^{ \infty } \frac{ {e}^{ - 2x} \: tanh(x)}{x} \: dx}$

$\displaystyle \tt \red{\int_{0}^{ \infty } {e}^{ - 2x}\frac{ \: 2 \sinh^{2} (x)}{2 \sinh(x) \cosh(x) } \: \frac{dx}{x}}$

$\displaystyle \tt \red{\int_{0}^{ \infty } {e}^{ - 2x}\frac{ \: \cosh^{} (2x) - 1}{\sinh(2x) } \: \frac{dx}{x}}$

$\displaystyle \tt \red{\int_{0}^{ \infty } {e}^{ - 2x}\frac{ \: e^{2x} + {e}^{ - 2x} - 2}{e^{2x} - {e}^{ - 2x} } \: \frac{dx}{x}}$

$\displaystyle \tt \red{\int_{0}^{ \infty } \frac{ \: e^{ - 2x} + {e}^{ - 6x} - 2 {e}^{ - 4x} }{^{} {1 - e}^{ - 4x} } \: \frac{dx}{x}}$

$\displaystyle \tt \red{\int_{1}^{ 0 } \frac{ {x}^{ - \frac{1}{2} } + {x}^{ \frac{1}{2} } - 2}{1 - x } \: \frac{1}{ \ln(x) } \: dx}$

$\displaystyle \tt \red{\int_{0}^{ \infty } \int_{0}^{ 1} \frac{ {x}^{t - \frac{1}{2} } + {x}^{ t + \frac{1}{2 {}^{} } } - 2 {x}^{t} }{1 - x } \: dxdt}$

$\displaystyle \tt \red {\int_{0}^{ \infty }\left ( - \int_{0}^{ 1 } \frac{1 - {x}^{t - \frac{1}{2} } }{1 - x} \: dx - \int_{0}^{ 1 } \frac{1 - {x}^{t + \frac{1}{2} } }{1 - x} \: dx + 2 \int_{0}^{ 1 } \frac{1 - {x}^{t} }{1 - x}dx \right)dt}$

$\displaystyle \bf \red{\int_{0}^{ \infty } \bigg[ - \Psi \bigg( t - \frac{1}{2} \bigg)- \Psi \bigg( t + \frac{3}{2} \bigg) +2 \Psi(t + 1) \bigg] \: dt }$

$\tt \red{ \ln \bigg( \dfrac{ \Gamma^{2}(t + 1) }{ \Gamma(t + \frac{1}{2} ) \Gamma(t + \frac{3}{2} )} \bigg) \bigg|_{t = 0}^{t \to \infty} }$

$\tt \red{ \ln(1) - \ln \bigg( \dfrac{ \Gamma^{2}( 1) }{ \Gamma(\frac{1}{2} ) \Gamma( \frac{3}{2} )} \bigg) }$

$\tt \red{ \ln\bigg( \dfrac{1}{2} {\Gamma}^{2} \bigg( \dfrac{1}{2} \bigg) \bigg) }$

$\red { \ln( \frac{\pi}{2} ) }$