"라마누잔의 정적분"의 두 판 사이의 차이

2012년 8월 25일 (토) 13:57 판

\(\int_{0}^{\infty}\frac{x e^{-\sqrt{5}x}}{\cosh{x}}\,dx=\frac{1}{8}(\psi^{(1)}(\frac{1+\sqrt{5}}{4})-\psi^{(1)}(\frac{3+\sqrt{5}}{4}))\)

Integrate[(x Exp[-x Sqrt[5]])/Cosh[x], {x, 0, \[Infinity]}] // FullSimplify

[%28x+Exp[-x+Sqrt[5]%29/Cosh[x],+%7Bx,+0,+[Infinity]%7D]+ http://www.wolframalpha.com/input/?i=Integrate[(x+Exp[-x+Sqrt[5]])/Cosh[x],+{x,+0,+[Infinity]}]+]

[1,%281%2Bsqrt%285%29%29/4-polygamma[1,%283%2Bsqrt%285%29%29/4]%29/8 http://www.wolframalpha.com/input/?i=(polygamma[1,(1%2Bsqrt(5))/4]-polygamma[1,(3%2Bsqrt(5))/4])/8]

\(\int_{0}^{\infty}\frac{x^{2}e^{-\sqrt{3}x}}{\sinh{x}}\,dx=-\frac{1}{4}\psi^{(2)}(\frac{1+\sqrt{3}}{4})\)

Integrate[(x^2 Exp[-x Sqrt[3]])/Sinh[x], {x, 0, \[Infinity]}] //FullSimplify

[%28x%5E2+Exp[-x+Sqrt[3]%29/Sinh[x],+%7Bx,+0,+Infinity%7D] http://www.wolframalpha.com/input/?i=integrate[(x^2+Exp[-x+Sqrt[3]])/Sinh[x],+{x,+0,+Infinity}]]

Berndt, B. C. and Rankin, R. A. Ramanujan: Letters and Commentary. Providence, RI: Amer. Math. Soc., 1995.

@@ 39번째 줄: / 39번째 줄: @@
 <h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">재미있는 사실</h5>
-* Math Overflow
+*   <br>
-* [http://mathoverflow.net/questions/66812/ramanujans-eccentric-integral-formula Ramanujan’s eccentric Integral formula]
+*  Math Overflow<br>
+** [http://mathoverflow.net/questions/66812/ramanujans-eccentric-integral-formula Ramanujan’s eccentric Integral formula]
-연분수 http://mathoverflow.net/questions/49866/applications-of-finite-continued-fractions/
+** 연분수 http://mathoverflow.net/questions/49866/applications-of-finite-continued-fractions/[http://mathoverflow.net/search?q= ]
-*
-* http://mathoverflow.net/search?q=
-* 네이버 지식인 http://kin.search.naver.com/search.naver?where=kin_qna&query=