對著只因哥猛猛地導(dǎo)

2023-03-28 21:54 作者:偏謬Lyx 0人讀過 | 我要投稿

本文介紹費曼積分法，即“在積分號下求導(dǎo)”。

Leibniz 公式

假設(shè)有如下形式的積分，

$I(t)%3D%5Cint_%7Ba(t)%7D%5E%7Bb(t)%7Df(x%2Ct)%5C%2C%5Cmathrm%7Bd%7Dx$

其中? $t$ ?為參數(shù)。由導(dǎo)數(shù)的定義可得，

$%5Cfrac%7B%5Cmathrm%7Bd%7DI%7D%7B%5Cmathrm%7Bd%7Dt%7D%3D%5Clim_%7B%5CDelta%7Bt%7D%5Cto0%7D%5Cfrac%7BI(t%2B%5CDelta%7Bt%7D)-I(t)%7D%7B%5CDelta%7Bt%7D%7D$

對于 $I(t%2B%5CDelta%7Bt%7D)$ ，其積分上下限為，

$%5Cint_%7Ba(t%2B%5CDelta%7Bt%7D)%7D%5E%7Bb(t%2B%5CDelta%7Bt%7D)%7D%3D%5Cint_%7Ba%2B%5CDelta%7Ba%7D%7D%5E%7Bb%2B%5CDelta%7Bb%7D%7D%3D%5Cint_%7Ba%2B%5CDelta%7Ba%7D%7D%5Ea%2B%5Cint_a%5Eb%2B%5Cint_b%5E%7Bb%2B%5CDelta%7Bb%7D%7D$

故函數(shù)增量為，

$%5Cbegin%7Bsplit%7D%0AI(t%2B%5CDelta%7Bt%7D)-I(t)%3D%26%5C%2C%5Cint_a%5Eb%5Cleft%5Bf(x%2Ct%2B%5CDelta%7Bt%7D)-f(x%2Ct)%5Cright%5D%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%5C%2C%2B%5Cint_b%5E%7Bb%2B%5CDelta%7Bb%7D%7Df(x%2Ct%2B%5CDelta%7Bt%7D)%5C%2C%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%5C%2C-%5Cint_a%5E%7Ba%2B%5CDelta%7Ba%7D%7Df(x%2Ct%2B%5CDelta%7Bt%7D)%5C%2C%5Cmathrm%7Bd%7Dx%0A%5Cend%7Bsplit%7D$

對于后兩項，當(dāng) $%5CDelta%7Bt%7D%5Cto0$ 時， $%5CDelta%7Ba%7D%2C%5CDelta%7Bb%7D%5Cto0$ ，于是有，

$%5Cbegin%7Balign%7D%0A%26%5Clim_%7B%5CDelta%7Bt%7D%5Cto0%7D%5Cint_b%5E%7Bb%2B%5CDelta%7Bb%7D%7Df(x%2Ct%2B%5CDelta%7Bt%7D)%5C%2C%5Cmathrm%7Bd%7Dx%3Df(b%2Ct)%5CDelta%7Bb%7D%5C%5C%0A%26%5Clim_%7B%5CDelta%7Bt%7D%5Cto0%7D%5Cint_a%5E%7Ba%2B%5CDelta%7Ba%7D%7Df(x%2Ct%2B%5CDelta%7Bt%7D)%5C%2C%5Cmathrm%7Bd%7Dx%3Df(a%2Ct)%5CDelta%7Ba%7D%0A%5Cend%7Balign%7D$

代回導(dǎo)數(shù)定義式可得，

$%5Cfrac%7B%5Cmathrm%7Bd%7DI%7D%7B%5Cmathrm%7Bd%7Dt%7D%3D%5Cint_a%5Eb%5Cfrac%7B%5Cpartial%7Bf(x%2Ct)%7D%7D%7B%5Cpartial%7Bt%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx%2Bf(b%2Ct)%5Cfrac%7B%5Cmathrm%7Bd%7Db%7D%7B%5Cmathrm%7Bd%7Dt%7D-f(a%2Ct)%5Cfrac%7B%5Cmathrm%7Bd%7Da%7D%7B%5Cmathrm%7Bd%7Dt%7D$

此即 Leibniz 公式。

Dirichlet 積分

計算積分，

$g(t)%3D%5Cint_0%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-xt%7D%5Cfrac%7B%5Csin(ax)%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%2C%5Cquad(t%3E0)$

求導(dǎo)可得，

$%5Cbegin%7Bsplit%7D%0A%5Cfrac%7B%5Cmathrm%7Bd%7Dg%7D%7B%5Cmathrm%7Bd%7Dt%7D%26%3D%5Cint_0%5E%5Cinfty-x%5C%2C%5Cmathrm%7Be%7D%5E%7B-xt%7D%5Cfrac%7B%5Csin(ax)%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3D-%5Cint_0%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-xt%7D%5Csin(ax)%5C%2C%5Cmathrm%7Bd%7Dx%0A%5Cend%7Bsplit%7D$

利用 Euler 公式? $%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dax%7D%3D%5Ccos(ax)%2B%5Cmathrm%7Bi%7D%5Csin(ax)$ ，可得

$%5Csin(ax)%3D%5Cfrac%7B1%7D%7B2%5Cmathrm%7Bi%7D%7D%5Cleft(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dax%7D-%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dax%7D%5Cright)$

于是積分化為，

$%5Cbegin%7Bsplit%7D%0A%5Cfrac%7B%5Cmathrm%7Bd%7Dg%7D%7B%5Cmathrm%7Bd%7Dt%7D%0A%26%3D-%5Cfrac%7B1%7D%7B2%5Cmathrm%7Bi%7D%7D%5Cint_0%5E%5Cinfty%5Cleft%5B%5Cmathrm%7Be%7D%5E%7B(%5Cmathrm%7Bi%7Da-t)x%7D-%5Cmathrm%7Be%7D%5E%7B-(%5Cmathrm%7Bi%7Da%2Bt)x%7D%5Cright%5D%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3D-%5Cfrac%7B1%7D%7B2%5Cmathrm%7Bi%7D%7D%5Cleft%5B%5Cfrac%7B%5Cmathrm%7Be%7D%5E%7B(%5Cmathrm%7Bi%7Da-t)x%7D%7D%7B%5Cmathrm%7Bi%7Da-t%7D%2B%5Cfrac%7B%5Cmathrm%7Be%7D%5E%7B-(%5Cmathrm%7Bi%7Da%2Bt)x%7D%7D%7B%5Cmathrm%7Bi%7Da%2Bt%7D%5Cright%5D_%7Bx%3D0%7D%5E%7Bx%3D%5Cinfty%7D%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B2%5Cmathrm%7Bi%7D%7D%5Cleft(%5Cfrac%7B1%7D%7B%5Cmathrm%7Bi%7Da-t%7D%2B%5Cfrac%7B1%7D%7B%5Cmathrm%7Bi%7Da%2Bt%7D%5Cright)%5C%5C%0A%26%3D-%5Cfrac%7Ba%7D%7Bt%5E2%2Ba%5E2%7D%0A%5Cend%7Bsplit%7D$

積分可得，

$g(t)%3DC-%5Carctan%5Cleft(%5Cfrac%7Bt%7D%7Ba%7D%5Cright)$

對于原積分表達(dá)式，有 $g(%5Cinfty)%3D0$ ，即

$C%3D%5Carctan(%5Cpm%5Cinfty)%3D%5Cpm%5Cfrac%7B%5Cpi%7D%7B2%7D$

其中正負(fù)號的選擇由? $a$ ?的正負(fù)決定。于是當(dāng)? $a%5Cneq0$ ?時，積分結(jié)果為，

$%5Cint_0%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-xt%7D%5Cfrac%7B%5Csin(ax)%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%3D%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cfrac%7Ba%7D%7B%7Ca%7C%7D-%5Carctan%5Cleft(%5Cfrac%7Bt%7D%7Ba%7D%5Cright)$

而當(dāng)? $a%3D0$ ?時，被積函數(shù)恒為零，積分結(jié)果顯然為零。

當(dāng) $t%3D0$ ?時，

$%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Csin(ax)%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%3D%5Cbegin%7Bcases%7D%0A%2B%5Cpi%2F2%2C%26(a%3E0)%5C%5C%0A0%2C%26(a%3D0)%5C%5C%0A-%5Cpi%2F2%2C%26(a%3C0)%0A%5Cend%7Bcases%7D$

Frullani 積分

計算積分，

$I(a%2Cb)%3D%5Cint_0%5E%5Cinfty%5Cfrac%7Bf(ax)-f(bx)%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%2C%5Cquad(a%2Cb%3E0)$

顯然，當(dāng)? $a%3Db$ ?時， $I%3D0$ 。

對? $a$ ?求導(dǎo)，

$%5Cbegin%7Bsplit%7D%0A%5Cfrac%7B%5Cpartial%7BI%7D%7D%7B%5Cpartial%7Ba%7D%7D%26%3D%5Cint_0%5E%7B%5Cinfty%7Df'(ax)%5C%2C%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3D%5Cfrac1a%5Cleft%5Bf(%5Cinfty)-f(0)%5Cright%5D%0A%5Cend%7Bsplit%7D$

對? $a$ ?積分可得，

$I(a%2Cb)%3D%5Cleft%5Bf(%5Cinfty)-f(0)%5Cright%5D%5Cln%7Ba%7D%2BC(b)$

其中? $C(b)$ ?為? $b$ ?的任意函數(shù)。由于該函數(shù)具有任意性，所以可以寫成更方便的形式，

$C(b)%5Cto%5Cleft%5Bf(%5Cinfty)-f(0)%5Cright%5D%5Cln%7BC(b)%7D$

于是積分化為，

$I(a%2Cb)%3D%5Cleft%5Bf(%5Cinfty)-f(0)%5Cright%5D%5Cln%5Cleft%5Ba%7B%5Ccdot%7DC(b)%5Cright%5D$

由于? $I(b%2Cb)%3D0$ ，可知? $b%7B%5Ccdot%7DC(b)%3D1$ ，即? $C(b)%3D%5Cfrac1b%E3%80%82$ 代回原式可得，

$%5Cint_0%5E%5Cinfty%5Cfrac%7Bf(ax)-f(bx)%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%3D%5Cleft%5Bf(%5Cinfty)-f(0)%5Cright%5D%5Cln%5Cleft(%5Cfrac%7Ba%7D%7Bb%7D%5Cright)$

? $f(x)%3D%5Carctan%7Bx%7D$

$%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Carctan(ax)-%5Carctan(bx)%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%3D%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cln%5Cleft(%5Cfrac%7Ba%7D%7Bb%7D%5Cright)$
$f(x)%3D%5Cmathrm%7Be%7D%5E%7B-x%7D$

$%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Cmathrm%7Be%7D%5E%7B-ax%7D-%5Cmathrm%7Be%7D%5E%7B-bx%7D%7D%7Bx%7D%5C%2C%5Cmathrm%7Bd%7Dx%3D-%5Cln%5Cleft(%5Cfrac%7Ba%7D%7Bb%7D%5Cright)$

Dini 積分

計算積分，

$I(%5Calpha)%3D%5Cint_0%5E%5Cpi%5Cln%5Cleft(1-2%5Calpha%5Ccos%7Bx%7D%2B%5Calpha%5E2%5Cright)%5Cmathrm%7Bd%7Dx$

對? $%5Calpha$ ?求導(dǎo)，

$%5Cbegin%7Bsplit%7D%0A%5Cfrac%7B%5Cmathrm%7Bd%7DI%7D%7B%5Cmathrm%7Bd%7D%5Calpha%7D%26%3D%5Cint_0%5E%5Cpi%5Cfrac%7B2%5Calpha-2%5Ccos%7Bx%7D%7D%7B1-2%5Calpha%5Ccos%7Bx%7D%2B%5Calpha%5E2%7D%5C%2C%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3D%5Cfrac1%5Calpha%5Cint_0%5E%5Cpi%5Cleft(1-%5Cfrac%7B1-%5Calpha%5E2%7D%7B1-2%5Calpha%5Ccos%7Bx%7D%2B%5Calpha%5E2%7D%5Cright)%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3D%5Cfrac%7B%5Cpi%7D%7B%5Calpha%7D-%5Cfrac1%5Calpha%5Cint_0%5E%5Cpi%5Cfrac%7B1-%5Calpha%5E2%7D%7B1-2%5Calpha%5Ccos%7Bx%7D%2B%5Calpha%5E2%7D%5C%2C%5Cmathrm%7Bd%7Dx%0A%5Cend%7Bsplit%7D$

令? $z%3D%5Ctan%5Cleft(%5Cfrac%7Bx%7D%7B2%7D%5Cright)$ ，使得，

$%5Ccos%7Bx%7D%3D%5Cfrac%7B1-z%5E2%7D%7B1%2Bz%5E2%7D$

$%5Cmathrm%7Bd%7Dz%3D%5Cfrac12%5Csec%5E2%5Cleft(%5Cfrac%7Bx%7D%7B2%7D%5Cright)%5Cmathrm%7Bd%7Dx%3D%5Cfrac%7B1%2Bz%5E2%7D%7B2%7D%5C%2C%5Cmathrm%7Bd%7Dx$

代回原式，積分變形為，

$%5Cfrac%7B%5Cmathrm%7Bd%7DI%7D%7B%5Cmathrm%7Bd%7D%5Calpha%7D%3D%5Cfrac%7B%5Cpi%7D%7B%5Calpha%7D-%5Cfrac%7B2%5Cbeta%7D%7B%5Calpha%7D%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Cmathrm%7Bd%7Dz%7D%7Bz%5E2%2B%5Cbeta%5E2%7D%2C%5Cquad%5Cleft(%5Cbeta%3D%5Cfrac%7B1-%5Calpha%7D%7B1%2B%5Calpha%7D%5Cright)$

利用積分公式，

$%5Cfrac%7B%5Cmathrm%7Bd%7DI%7D%7B%5Cmathrm%7Bd%7D%5Calpha%7D%3D%5Cfrac%7B%5Cpi%7D%7B%5Calpha%7D-%5Cfrac%7B2%7D%7B%5Calpha%7D%5Carctan%5Cleft(%5Cfrac%7Bz%7D%7B%5Cbeta%7D%5Cright)%5CBigg%7C_0%5E%5Cinfty$

當(dāng)? $%5Calpha%3E1$ ?時， $%5Cbeta%3C0$ ，

$%5Cfrac%7B%5Cmathrm%7Bd%7DI%7D%7B%5Cmathrm%7Bd%7D%5Calpha%7D%3D%5Cfrac%7B%5Cpi%7D%7B%5Calpha%7D-%5Cfrac%7B2%7D%7B%5Calpha%7D%5Carctan(-%5Cinfty)%3D%5Cfrac%7B2%5Cpi%7D%7B%5Calpha%7D$

當(dāng)? $0%5Cle%5Calpha%3C1$ ?時， $%5Cbeta%3E0$ ，

$%5Cfrac%7B%5Cmathrm%7Bd%7DI%7D%7B%5Cmathrm%7Bd%7D%5Calpha%7D%3D%5Cfrac%7B%5Cpi%7D%7B%5Calpha%7D-%5Cfrac%7B2%7D%7B%5Calpha%7D%5Carctan(%5Cinfty)%3D0$

由于

$I(0)%3D%5Cint_0%5E%5Cpi%5Cln(1)%5C%2C%5Cmathrm%7Bd%7D%5Calpha%3D0$

所以，

$I(%5Calpha)%3D0%2C%5Cquad%5Cleft(0%5Cle%5Calpha%3C1%5Cright)$

對于? $%5Calpha%3E1$ ，有

$%5Cbegin%7Bsplit%7D%0A0%3DI%5Cleft(%5Cfrac1%5Calpha%5Cright)%26%3D%5Cint_0%5E%5Cpi%5Cln%5Cleft(1-%5Cfrac2%5Calpha%5Ccos%7Bx%7D%2B%5Cfrac%7B1%7D%7B%5Calpha%5E2%7D%5Cright)%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3D%5Cint_0%5E%5Cpi%5Cln%5Cleft(%5Cfrac%7B%5Calpha%5E2-2%5Calpha%5Ccos%7Bx%7D%2B1%7D%7B%5Calpha%5E2%7D%5Cright)%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3DI(%5Calpha)-%5Cint_0%5E%5Cpi%5Cln%5Calpha%5E2%5C%2C%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3DI(%5Calpha)-2%5Cpi%5Cln%5Calpha%0A%5Cend%7Bsplit%7D$

積分結(jié)果為，

$%5Cint_0%5E%5Cpi%5Cln%5Cleft(1-2%5Calpha%5Ccos%7Bx%7D%2B%5Calpha%5E2%5Cright)%5Cmathrm%7Bd%7Dx%3D%5Cbegin%7Bcases%7D%0A0%2C%260%5Cle%5Calpha%3C1%5C%5C%0A2%5Cpi%5Cln%5Calpha%2C%26%5Calpha%3E1%0A%5Cend%7Bcases%7D$

標(biāo)簽：數(shù)學(xué)積分求導(dǎo)費曼積分

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對著只因哥猛猛地導(dǎo)

Leibniz 公式

Dirichlet 積分

Frullani 積分

Dini 積分