寫幾個積分

2022-07-24 18:32 作者:子瞻Louis 0人讀過 | 我要投稿

好久沒更新了，今天就隨便寫幾道積分叭（可以拿去考考你的朋友，嘻嘻~）。

本文的題目都是up隨便找的，其中有的是經(jīng)過up的人性化改編。解決積分時會出現(xiàn)一些交換積分和求和次序的地方，這里并沒有詳細證明可以它們交換，若有讀者可以自行驗證（其實是我懶）

Q1： $I_1%3D%5Cint_0%5E1%5Cfrac%7B%5Csqrt%5B3%5D%7Bx(1-x)%5E2%7D%7D%7B(1%2Bx)%5E3%7D%5Cmathrm%20dx$

A：首先

$I_1%3D%5Cint_0%5E1%5Cleft(%5Cfrac%20x%7B1%2Bx%7D%5Cright)%5E%7B1%2F3%7D%5Cleft(%5Cfrac%20%7B1-x%7D%7B1%2Bx%7D%5Cright)%5E%7B2%2F3%7D%5Ccdot%5Cfrac%7B%5Cmathrm%20dx%7D%7B(1%2Bx)%5E2%7D$

作代換? $t%3D2x%2F(1%2Bx)$ ，得

$%5Cbegin%7Balign%7DI_1%26%3D%5Cfrac1%7B2%5Csqrt%5B3%5D2%7D%5Cint_0%5E%7B1%7Dt%5E%7B1%2F3%7D(1-t)%5E%7B2%2F3%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac1%7B2%5Csqrt%5B3%5D2%7DB%5Cleft(%5Cfrac43%2C%5Cfrac53%5Cright)%5C%5C%26%3D%5Cfrac%7B1%7D%7B4%5Csqrt%5B3%5D2%7D%5CGamma%5Cleft(%5Cfrac43%5Cright)%5CGamma%5Cleft(%5Cfrac53%5Cright)%5C%5C%26%3D%5Cfrac1%7B4%5Csqrt%5B3%5D2%7D%5Ccdot%5Cfrac13%5Ccdot%5Cfrac23%5Ccdot%5CGamma%5Cleft(%5Cfrac13%5Cright)%5CGamma%5Cleft(%5Cfrac23%5Cright)%5C%5C%26%3D%5Cfrac%7B%5Cpi%7D%7B9%5Csqrt%5B3%5D2%5Ccdot%5Csqrt3%7D%3D%5Cfrac%5Cpi%7B9%5Ccdot%7B108%7D%5E%7B1%2F6%7D%7D%5Cend%7Balign%7D$

其中? $B$ ?是beta函數(shù)，倒數(shù)第二等號是由于Gamma函數(shù)的遞推性質(zhì)，最后一個等號則是余元公式。

$%5Csquare$

Q2： $I_2%3D%5Cint_0%5E%5Cinfty%20%5Cfrac%7B%5Csqrt%20x%5Cln%5E2%20x%7D%7B1%2Bx%5E2%7D%5Cmathrm%20dx$

A：置

$P(z)%3A%3D%5Cint_0%5E%5Cinfty%5Cfrac%7Bx%5E%7Bz%7D%7D%7B1%2Bx%5E2%7D%5Cmathrm%20dx$

那么? $I_2%3D%5Cfrac%7B%5Cpartial%5E2P%7D%7B%5Cpartial%20z%5E2%7D%5Cleft(%5Cfrac12%5Cright)$

令? $%5Ctheta%3D%5Carctan%20x$ ，則

$%5Cbegin%7Balign%7DP(z)%26%3D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Ctan%5E%7Bz%7D%5Ctheta%20%5Cmathrm%20d%5Ctheta%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B%5Cpi%2F2%7D%5Csin%5Ez%5Ctheta%5Ccos%5E%7B-z%7D%5Ctheta%5Cmathrm%20dz%5C%5C%26%3D%5Cfrac12B%5Cleft(%5Cfrac%7B1%2Bz%7D2%2C%5Cfrac%7B1-z%7D2%5Cright)%5C%5C%26%3D%5Cfrac12%5Ccdot%5Cfrac%5Cpi%7B%5Ccos%5Cfrac%7B%5Cpi%20z%7D2%7D%5Cend%7Balign%7D$

最后一個等號用了余元公式。從而

$I_2%3D%5Cleft.%5Cfrac%5Cpi2%5Ccdot%5Cfrac%7B%5Cmathrm%20d%5E2%7D%7B%5Cmathrm%20dz%5E2%7D%5Cfrac%7B1%7D%7B%5Ccos%5Cfrac%7B%5Cpi%20z%7D2%7D%5Cright%7C_%7Bz%3D%5Cfrac12%7D%3D%5Cfrac%7B3%5Cpi%5E3%7D%7B2%5E%7B5%2F2%7D%7D$

$%5Csquare$

Q3： $I_3%3D%5Cint_0%5E1%5Cfrac%7B(1-x)%5Cln%5E2%20x%7D%7B1%2Bx%5E3%7D%5Cmathrm%20dx$

A：直接暴力展開

$%5Cbegin%7Balign%7DI_3%26%3D%5Cint_0%5E1(1-x)%5Cln%5E2x%5Csum_%7Bn%3D0%7D%5E%5Cinfty(-1)%5Enx%5E%7B3n%7D%5Cmathrm%20dx%5C%5C%26%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty(-1)%5En%5Cint_0%5E1x%5E%7B3n%7D(1-x)%5Cln%5E2x%5Cmathrm%20dx%5C%5C%26%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty(-1)%5En%5Cleft.%5Cfrac%7B%5Cpartial%5E2%7D%7B%5Cpartial%20%5Calpha%5E2%7DB(%5Calpha%2C%5Cbeta)%5Cright%7C_%7B%5Calpha%3D3n%2B1%2C%5Cbeta%3D2%7D%5C%5C%26%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty(-1)%5En%20%5Cfrac%7B%5Cmathrm%20d%5E2%7D%7B%5Cmathrm%20d%5Calpha%5E2%7D%5Cleft.%5Cfrac%7B%5CGamma(%5Calpha)%7D%7B%5CGamma(%5Calpha%2B2)%7D%5Cright%7C_%7B%5Calpha%3D3n%2B1%7D%5C%5C%26%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty(-1)%5En%5Cleft%5B%5Cfrac2%7B(3n%2B1)%5E3%7D-%5Cfrac2%7B(3n%2B2)%5E3%7D%5Cright%5D%5Cend%7Balign%7D$

令? $%5Comega%3De%5E%7B2i%5Cpi%2F3%7D$ ?，以及

$Q(t)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B(-1)%5E%7Bn%7D%7D%7Bn%5E3%7Dt%5E%7Bn%7D$

那么當? $k%3D1%2C2$ ?時

$%5Cbegin%7Balign%7D%26%5Cfrac13(Q(t)%2B%5Comega%5E%7B-k%7DQ(%5Comega%20t)%2B%5Comega%5E%7B-2k%7DQ(%5Comega%5E2%20t))%5C%5C%26%3D%5Cfrac13%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B(-1)%5En%7D%7Bn%5E3%7D(1%2B%5Comega%5E%7Bn-k%7D%2B%5Comega%5E%7B2n-2k%7D)t%5En%5C%5C%26%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20%5Cfrac%7B(-1)%5E%7B3n%2Bk%7D%7D%7B(3n%2Bk)%5E3%7Dt%5E%7B3n%2Bk%7D%5Cend%7Balign%7D$

取? $t%3D1%2Ck%3D1%2C2$ ?即可得

$%5Cbegin%7Balign%7DI_2%26%3D-%5Cfrac23(Q(1)%2B%5Comega%5E%7B-1%7DQ(%5Comega)%2B%5Comega%5E%7B-2%7DQ(%5Comega%5E2))%5C%5C%26-%5Cfrac23(Q(1)%2B%5Comega%5E%7B-2%7DQ(%5Comega)%2B%5Comega%5E%7B-4%7DQ(%5Comega%5E2))%5C%5C%26%3D-%5Cfrac43Q(1)-%5Cfrac23Q(1)%2B%5Cfrac23(Q(1)%2BQ(%5Comega)%2BQ(%5Comega%5E2))%0A%5C%5C%26%3D-2Q(1)%2B%5Cfrac2%7B27%7DQ(1)%5C%5C%26%3D%5Cfrac32%5Czeta(3)-%5Cfrac1%7B18%7D%5Czeta(3)%3D%5Cfrac%7B13%7D9%5Czeta(3)%26%5Cend%7Balign%7D$

其中? $%5Czeta(3)$ ?是Riemann?zeta函數(shù)取3的函數(shù)值，它也叫做Apéry常數(shù)。

有的積分算著算著就算成了級數(shù)

$%5Csquare$

Q4： $I_4%3D%5Cint_0%5E1%20%5Cfrac%7Bx(1-x)%5E2%7D%7B%5Csqrt%7B2-x%7D%7D%5Cmathrm%20dx$

A：依然考慮暴力展開

$%5Cbegin%7Balign%7DI_4%26%3D%5Cfrac1%7B%5Csqrt2%7D%5Cint_0%5E1%7Bx(1-x)%5E2%7D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cbinom%7B2n%7D%7Bn%7D%5Cfrac%7Bx%5En%7D%7B8%5En%7D%5Cmathrm%20dx%5C%5C%26%3D%5Cfrac1%7B%5Csqrt2%7D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cbinom%7B2n%7D%7Bn%7D%5Cfrac1%7B8%5En%7D%5Cint_0%5E1%7Bx%5E%7Bn%2B1%7D(1-x)%5E2%7D%5Cmathrm%20dx%5C%5C%26%3D%5Cfrac1%7B%5Csqrt2%7D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cbinom%7B2n%7Dn%5Cfrac1%7B8%5En%7DB(n%2B2%2C3)%5C%5C%26%3D%5Csqrt2%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cbinom%7B2n%7Dn%5Cfrac%7B1%7D%7B8%5En%7D%5Ccdot%5Cfrac1%7B(n%2B2)(n%2B3)(n%2B4)%7D%5Cend%7Balign%7D$

置

$F(z)%3A%3D%7B%5Csqrt2%7D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cbinom%7B2n%7Dn%5Cfrac%7B1%7D%7B8%5En%7D%5Cfrac%7Bx%5E%7Bn%2B4%7D%7D%7B(n%2B2)(n%2B3)(n%2B4)%7D$

則? $I_4%3DF(1)%2CF'''(x)%3D2x%2F%5Csqrt%7B2-x%7D$ ，所以

$%5Cbegin%7Balign%7D%0AI_4%26%3D%5Cint_0%5E1%5Cint_%7B0%7D%5E%7Bz%7D%5Cint_0%5E%7By%7D%5Cfrac%20%7B2x%7D%7B%5Csqrt%20%7B2-x%7D%7D%5Cmathrm%20dx%5Cmathrm%20dy%5Cmathrm%20dz%0A%5C%5C%26%3D-%5Cfrac43%5Cint_0%5E1%5Cint_0%5Ez%5Csqrt%7B2-y%7D%5Cleft(y%2B4%5Cright)%5Cmathrm%20dy%5Cmathrm%20dz%2B%5Cfrac%7B2%5E%7B7%2F2%7D%7D%7B3%7D%0A%5C%5C%26%3D-%5Cfrac43%5Cint_0%5E1%5Cint_y%5E1%5Csqrt%7B2-y%7D%5Cleft(y%2B4%5Cright)%5Cmathrm%20dz%5Cmathrm%20dy%2B%5Cfrac%7B2%5E%7B7%2F2%7D%7D%7B3%7D%0A%5C%5C%26%3D%5Cfrac43%5Cint_0%5E1%5Csqrt%7B2-y%7D%5Cleft(y%2B4%5Cright)(y-1)%5Cmathrm%20dy%2B%5Cfrac%7B2%5E%7B7%2F2%7D%7D%7B3%7D%0A%5C%5C%26%3D%5Cfrac%7B2%5E%7B7%2F2%7D%7D%7B3%7D-%5Cfrac%20%7B16%7D%7B105%7D(13%2B8%5Csqrt2)%0A%5C%5C%26%3D%5Cfrac8%7B105%7D(19%5Csqrt2-26)%0A%5Cend%7Balign%7D$ ?

$%5Csquare$

沒有積分遼，所以最后來道級數(shù)收尾吧

Q5： $I_5%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B1%7D%7B(n%5E2%2B1)%5E2%7D$

A：考慮余切函數(shù)的Fourier展開

$%5Ccot%5Cpi%20z%3D%5Cfrac1%7B%5Cpi%20z%7D%2B%5Cfrac1%5Cpi%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B2z%7D%7Bz%5E2-n%5E2%7D$

令? $z%3Di$ ?，得

$i%5Cfrac%7Be%5E%7B2%5Cpi%7D%2B1%7D%7Be%5E%7B2%5Cpi%7D-1%7D%3D%5Cfrac1%7B%5Cpi%20i%7D-%5Cfrac1%5Cpi%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B2i%7D%7Bn%5E2%2B1%7D$

也就是

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5E2%2B1%7D%3D%5Cfrac%5Cpi2%5Ccdot%5Cfrac%7Be%5E%7B2%5Cpi%7D%2B1%7D%7Be%5E%7B2%5Cpi%7D-1%7D-%5Cfrac12$

對余切函數(shù)取一次導數(shù)并令? $z%3Di$ 得到

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B1%7D%7B(n%5E2%2B1)%5E2%7D%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bn%5E2%7D%7B(n%5E2%2B1)%5E2%7D%2B%5Cfrac%7B2e%5E%7B2%5Cpi%7D%7D%7B(e%5E%7B2%5Cpi%7D-1)%5E2%7D-%5Cfrac1%7B2%5Cpi%7D$

而又有

$%5Cbegin%7Balign%7D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bn%5E2%7D%7B(n%5E2%2B1)%5E2%7D%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5E2%2B1%7D-%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B1%7D%7B(n%5E2%2B1)%5E2%7D%5C%5C%26%3D%5Cfrac%5Cpi2%5Ccdot%5Cfrac%7Be%5E%7B2%5Cpi%7D%2B1%7D%7Be%5E%7B2%5Cpi%7D-1%7D-%5Cfrac12-%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B1%7D%7B(n%5E2%2B1)%5E2%7D%5Cend%7Balign%7D$

所以

$I_5%3D%5Cfrac%5Cpi4%5Ccdot%5Cfrac%7Be%5E%7B2%5Cpi%7D%2B1%7D%7Be%5E%7B2%5Cpi%7D-1%7D%2B%5Cfrac%7Be%5E%7B2%5Cpi%7D%7D%7B(e%5E%7B2%5Cpi%7D-1)%5E2%7D-%5Cfrac%5Cpi4-%5Cfrac14$

$%5Csquare$

一杯茶，一包煙，一道積分寫一天。

E.N.D

標簽：

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