Deus ex machina (Latin: [ˈdeus eks ˈmaː.kʰi.naː]: /ˈdeɪ.əs ɛks ˈmɑːkiːnə/ or /ˈdiːəs ɛks ˈmækɨnə/; plural: dei ex machina) from Latin deus, meaning "a god", ex, meaning "from", and machina, meaning "a device, a scaffolding, an artifice", is a calque from Greek ἀπὸ μηχανῆς θεός (apò mēkhanḗs theós), meaning "god from the machine". The term has evolved into a plot device whereby a seemingly unsolvable problem is suddenly and abruptly resolved by the contrived and unexpected intervention of some new event, character, ability or object. Depending on how it is done, it can be intended to move the story...
Nowadays I encounter a integral which is difficult for me to solve it,Please help me to find it.Thank you!
$$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{C}{2}-\ln2$$
Where C is EulerGamma Constant
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Hint: make use of the Binet's second formula http://mathworld.wolfram.com/BinetsLogGammaFormulas.html.
$$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx$$
$$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-2\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{4\pi x}-1)} dx$$
$$=\int_{0}^{\infty}\frac{x...
0
Hint: make use of the Binet's second formula http://mathworld.wolfram.com/BinetsLogGammaFormulas.html.
$$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx$$
$$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-2\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{4\pi x}-1)} dx$$
$$=\int_{0}^{\infty}\frac{x...
