Chapter 47: Feynman method

47.1 Feynman method of differentiation / derivative technique

https://www.bilibili.com/video/BV1hG411Z7Cb

分式微分不是難而是煩

47.1.1 principle

Theorem 25.1 Feynman method of differentiation / derivative

f(t)=k[u(t)]a[v(t)]b[w(t)]c⋯⇓f′(t)=f(t)[au′(t)u(t)+bv′(t)v(t)+cw′(t)w(t)+⋯]

Proof:

f(t)=k[u(t)]a[v(t)]b[w(t)]c⋯

f=kuavbwc⋯=k⋅ua⋅vb⋅wc⋅⋯

f=kuavbwc⋯=k⋅ua⋅vb⋅wc⋅⋯lnf=ln(kuavbwc⋯)=lnk+lnua+lnvb+lnwc+⋯=lnk+alnu+blnv+clnw+⋯ddtlnf=ddt(lnk+alnu+blnv+clnw+⋯)ddtff=0+ddt(alnu)+ddt(blnv)+ddt(clnw)+⋯=addtlnu+bddtlnv+cddtlnw+⋯=addtuu+bddtvv+cddtww+⋯f′f=au′u+bv′v+cw′w+⋯f′=f(au′u+bv′v+cw′w+⋯)f′(t)=f(t)[au′(t)u(t)+bv′(t)v(t)+cw′(t)w(t)+⋯]

47.1.2 examples

f(x)=xx

(xx)′=xx+xxlnx

f(x)=xxlnf(x)=xlnxddxlnf(x)=ddx[xlnx]f′(x)f(x)=[xlnx]′f′(x)=f(x)[xlnx]′=xx[(x)′lnx+x(lnx)′]=xx[1lnx+x1x]=xx[lnx+1]=xx[1+lnx]=xx+xxlnx

3D delta function[48.2.2]

Δ(1r)=∇2(1√r2)=∇⋅∇(1√r⋅r)=∇⋅(∂∂x,∂∂y,∂∂z)(1√x2+y2+z2)

∂∂x1√x2+y2+z2

∂∂x1√x2+y2+z2=∂∂x[(x2+y2+z2)−12]=(x2+y2+z2)−12[−122xx2+y2+z2]=−x(x2+y2+z2)−32

∇(1r)=∇(1√r⋅r)=(∂∂x,∂∂y,∂∂z)(1√x2+y2+z2)=(−x(x2+y2+z2)32,−y(x2+y2+z2)32,−z(x2+y2+z2)32)

∇⋅∇(1r)=∇⋅(−x(x2+y2+z2)32,−y(x2+y2+z2)32,−z(x2+y2+z2)32)=∂∂x−x(x2+y2+z2+ϵ2)32+∂∂y−y(x2+y2+z2+ϵ2)32+∂∂z−z(x2+y2+z2+ϵ2)32

∂∂x−x(x2+y2+z2+ϵ2)32

∂∂x−x(x2+y2+z2)32=−x(x2+y2+z2)32[1⋅1x+−322xx2+y2+z2]=−1(x2+y2+z2)32+3x2(x2+y2+z2)52=−(x2+y2+z2)+3x2(x2+y2+z2)52=2x2−y2−z2(x2+y2+z2)52

∇⋅∇(1r)=∂∂x−x(x2+y2+z2+ϵ2)32+∂∂y−y(x2+y2+z2+ϵ2)32+∂∂z−z(x2+y2+z2+ϵ2)32=+2x2−y2−z2(x2+y2+z2)52+−x2+2y2−z2(x2+y2+z2)52+−x2−y2+2z2(x2+y2+z2)52=0

Δ(1r)=∇⋅∇(1r)=0

6(1+2t2)(t3−t)2√t+5t2(4t)32+√1+2tt+√1+t2

ddt[6(1+2t2)(t3−t)2√t+5t2(4t)32+√1+2tt+√1+t2]

ddt[6(1+2t2)(t3−t)2√t+5t2(4t)32]

ddt[6(1+2t2)(t3−t)2√t+5t2(4t)32]=6(1+2t2)(t3−t)2√t+5t2(4t)32⋅[=6[1+2t2](t3−t)2√t+5t2(4t)32⋅[1⋅11+2t2⋅4t,[1+2t2]→{exponential:1linear to denominator:11+2t2differentiation4t=6(1+2t2)[(t3−t)2]√t+5t2(4t)32⋅[4t1+2t2+2⋅1t3−t⋅(3t2−1),[(t3−t)2]→{exponential:2linear to denominator:1t3−tdifferentiation3t2−1=6(1+2t2)(t3−t)2[√t+5t2](4t)32⋅[4t1+2t2+6t2−2t3−t+−12⋅1t+5t2⋅(1+10t),[√t+5t2]→{exponential:−12linear to denominator:1t+5t2differentiation1+10t=6(1+2t2)(t3−t)2√t+5t2[(4t)32]⋅[4t1+2t2+6t2−2t3−t−1+10t2t+10t2+−32⋅14t⋅4],[(4t)32]→{exponential:−32linear to denominator:14tdifferentiation4=6(1+2t2)(t3−t)2√t+5t2(4t)32[4t1+2t2+6t2−2t3−t−1+10t2t+10t2−32t]

ddt[√1+2tt+√1+t2]

ddt[√1+2tt+√1+t2]=√1+2tt+[√1+t2][1221+2t+(−1)1+[122t√1+t2]t+√1+t2]

47.2 Feynman method of integration / integral technique

47.2.1 principle

https://www.bilibili.com/video/BV1Lj411L79X/?t=2m38s

https://www.youtube.com/watch?v=GW86SShcYbM

I=∫f(x)dx

I(t)=∫f(x,t)dx

I=∫f(x)dxI(t)=∫f(x,t)dxddtI(t)=I′(t)=ddt∫f(x,t)dxLebniz integral rule=f(x,b(x),t)db(x)dx−f(x,a(x),t)da(x)dx+∫b(x)a(x)∂∂tf(x,t)dx⋯=∫∂∂tf(x,t)dx=∫∂f(x,t)∂tdx=∫ft(x,t)dxI(t)=∫I′(t)dt=∫I′(t)dt+C=[∫I′(t)dt](t)⇓ if f(x,t=0)=f(x,0)=f(x)If(x,0)=f(x)=I(0)=[∫I′(t)dt](0)

47.2.2 Dirichlet integral

as a example by Feynman method of integration / integral technique

https://en.wikipedia.org/wiki/Dirichlet_integral

https://www.bilibili.com/video/BV1Lj411L79X/?t=4m38s

∫∞0sinxxdx

https://www.youtube.com/watch?v=ZZccxuOpb4k

https://blog.csdn.net/zhuoqingjoking97298/article/details/127950915

∫∞−∞sinxxdx

∫∞−∞sinxxdx

f(x)=sinxx is an even function⇔f(−x)=sin(−x)(−x)=−sin(x)−x=sinxx=f(x)

∫∞−∞sinxxdx=∫0−∞sinxxdx+∫∞0sinxxdx=∫x=0x=−∞sinxxdx+∫∞0sinxxdx=∫x=0x=−∞sinxxdx(,x′=−x⇔x=−x′)+∫∞0sinxxdx=∫(−x′)=0(−x′)=−∞sin(−x′)(−x′)d(−x′)+∫∞0sinxxdx=∫x′=−0x′=−(−∞)−sinx′−x′(−dx′)+∫∞0sinxxdx=∫x′=0x′=∞sinx′x′(−dx′)+∫∞0sinxxdx=−∫x′=0x′=∞sinx′x′dx′+∫∞0sinxxdx=∫x′=∞x′=0sinx′x′dx′+∫∞0sinxxdx=∫∞0sinxxdx+∫∞0sinxxdx=2∫∞0sinxxdx

∫∞−∞sinxxdx=2∫∞0sinxxdx

∫∞0sinxxdx

I=∫∞0sinxxdx=∫∞0f(x)dx,f(x)=sinxx∫∞0e−txsinxxdx=∫∞0e−txf(x)dx, Laplacian transform of f(x)=∫∞0f(x,t)dx,{f(x,t)=e−txf(x)f(x,0)=e−0xf(x)=1⋅f(x)=f(x)⇓I(t)=∫∞0f(x,t)dx=∫∞0e−txf(x)dx=∫∞0e−txsinxxdx⇒I(0)=IddtI(t)=I′(t)=ddt∫∞0f(x,t)dx=∫∞0∂f(x,t)∂tdx=∫∞0∂e−txsinxx∂tdx=∫∞0sinxx[∂e−tx∂t]dx=∫∞0sinxx[−xe−tx]dx=∫∞0sinx[−e−tx]dx=−∫∞0e−txsinxdx, Laplacian transform of −sinx=∫∞0e−tx(−sinx)dx=∫x=∞x=0e−txdcosx,I′(t)=∫∞0e−tx(−sinx)dx=[e−txcosx]x=∞x=0−∫x=∞x=0cosxde−tx=[e−t∞cos∞]−[e−t0cos0]−∫x=∞x=0cosx(−te−txdx)=[0cos∞]−[1⋅1]+∫∞0te−txcosxdx=0−1+∫∞0te−txcosxdx=−1+∫x=∞x=0te−txdsinx=−1+[te−txsinx]x=∞x=0−∫x=∞x=0sinxd(te−tx)=−1+[0−0]−∫∞0sinx(te−tx(−t)dx)=−1−∫∞0t2e−tx(−sinx)dx=−1−t2∫∞0e−tx(−sinx)dx=−1−t2I′(t)I′(t)=−1−t2I′(t)I′(t)=−11+t2I(t)=∫−11+t2dtt=tanθ=∫−11+tan2θdtanθ=−∫1sec2θsec2θdθ=−∫dθ=−θ+C=−arctant+C=−tan−1t+C∫∞0e−txsinxxdx=I(t)=−arctant+C⇓⇒I=∫∞0sinxxdx=∫∞0e−0xsinxxdx=I(0)=−arctan0+C=−0+C⇒C=I⇒0=∫∞00sinxxdx=∫∞0e−∞xsinxxdx=I(∞)=−arctan∞+C=−π2+C0=−π2+C⇒C=π2I=I(0)=C=π2,I=∫∞0sinxxdx∫∞0sinxxdx=π2

∫∞0sinxxdx=π2

∫∞−∞sinxxdx=2∫∞0sinxxdx=2⋅π2=π

∫∞−∞sinxxdx=π

47.2.3 Gaussian integral

https://www.youtube.com/watch?v=jP-6j6mEpRg

∫∞−∞e−x2dx

∫∞−∞e−x2dx

f(x)=e−x2 is an even function⇔f(−x)=e−(−x)2=e−x2=f(x)

∫∞−∞e−x2dx=∫0−∞e−x2dx+∫∞0e−x2dx=∫x=0x=−∞e−x2dx+∫∞0e−x2dx=∫x=0x=−∞e−x2dx(,x′=−x⇔x=−x′)+∫∞0e−x2dx=∫(−x′)=0(−x′)=−∞e−(−x′)2d(−x′)+∫∞0e−x2dx=∫x′=−0x′=−(−∞)e−(x′)2(−dx′)+∫∞0e−x2dx=∫x′=0x′=∞e−(x′)2dx′+∫∞0e−x2dx=−∫x′=0x′=∞e−(x′)2dx′+∫∞0e−x2dx=∫x′=∞x′=0e−(x′)2dx+∫∞0e−x2dx=∫∞0e−x2dx+∫∞0e−x2dx=2∫∞0e−x2dx

∫∞−∞e−x2dx=2∫∞0e−x2dx

∫∞0e−x2dx

I(t)=[∫t0e−x2dx]2

I′(t)=dI(t)dt=d[∫t0e−x2dx]2dt=d[∫t0e−x2dx]2d[∫t0e−x2dx]d[∫t0e−x2dx]dt=2[∫t0e−x2dx][ddt∫t0e−x2dx]=2[∫t0e−x2dx][ddt∫t0e−x2dx]FToC=2[∫t0e−x2dx][e−t2]=2∫t0e−x2e−t2dx=2∫t0e−(x2+t2)dx=2∫x=tx=0e−(x2+t2)dx,x′=xt⇔x=tx′=2∫tx′=ttx′=0e−((tx′)2+t2)dtx′=2∫x′=1x′=0e−t2((x′)2+1)tdx′=2∫10te−t2(x2+1)dx=−∫10(−2t)e−t2(x2+1)dx=−∫10∂∂te−t2(x2+1)x2+1dx=−ddt∫10e−t2(x2+1)x2+1dx[∫t0e−x2dx]2=I(t)=−∫10e−t2(x2+1)x2+1dx+C0=[∫00e−x2dx]2=I(0)=∫101x2+1dx+Cx=tanθ=[−arctanx]10+C=−π4+C⇒C=π4[∫t0e−x2dx]2=I(t)=−∫10e−t2(x2+1)x2+1dx+C=−∫10e−t2(x2+1)x2+1dx+π4[∫∞0e−x2dx]2=lim

\int_{0}^{\infty}\mathrm{e}^{-x^{2}}\mathrm{d}x=\dfrac{\sqrt{\pi}}{2}

\int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}}\mathrm{d}x=2\int_{0}^{\infty}\mathrm{e}^{-x^{2}}\mathrm{d}x=2\cdot\dfrac{\sqrt{\pi}}{2}=\sqrt{\pi}

\int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}}\mathrm{d}x=\sqrt{\pi}

\tag*{$\Box$}

47.2.4 other examples

https://www.youtube.com/watch?v=GW86SShcYbM

\int_{0}^{1}\ln x\mathrm{d}x

I=\int_{0}^{1}\ln x\mathrm{d}x=\int_{0}^{1}\ln\left(x\right)\mathrm{d}x,f\left(x\right)=\ln x

I\left(t\right)=\int_{0}^{1}\ln tx\mathrm{d}x=\int_{0}^{1}f\left(tx\right)\mathrm{d}x=\int_{0}^{1}f\left(x,t\right)\mathrm{d}x,\begin{cases} f\left(x,t\right)=f\left(tx\right)=\ln tx\\ f\left(x,1\right)=f\left(x\right)=\ln x \end{cases}

\begin{aligned} I\left(t\right)= & \int_{0}^{1}f\left(x,t\right)\mathrm{d}x=\int_{0}^{1}f\left(tx\right)\mathrm{d}x=\int_{0}^{1}\ln tx\mathrm{d}x\Rightarrow I\left(1\right)=I\\ \dfrac{\mathrm{d}}{\mathrm{d}t}I\left(t\right)=I^{\prime}\left(t\right)= & \dfrac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}f\left(x,t\right)\mathrm{d}x=\int_{0}^{1}\dfrac{\partial f\left(x,t\right)}{\partial t}\mathrm{d}x=\int_{0}^{1}\dfrac{\partial\ln tx}{\partial t}\mathrm{d}x\\ = & \int_{0}^{1}\dfrac{1}{tx}x\mathrm{d}x=\int_{0}^{1}\dfrac{1}{t}\mathrm{d}x=\dfrac{1}{t}\int_{0}^{1}\mathrm{d}x=\dfrac{1}{t}\left[x\right]_{0}^{1}=\dfrac{1}{t}\\ I\left(t\right)= & \int I^{\prime}\left(t\right)\mathrm{d}t=\int\dfrac{1}{t}\mathrm{d}t=\ln\left|t\right|+C\\ \Downarrow & \Rightarrow I=\int_{0}^{1}\ln x\mathrm{d}x=\int_{0}^{1}\ln1x\mathrm{d}x=I\left(1\right)=\ln\left|1\right|+C=0+C\Rightarrow C=I\\ \text{no more known } & \text{boundary condition} \end{aligned}

\begin{aligned} \Im\left(t\right)= & \int_{0}^{1}x^{t}\mathrm{d}x=\left[\dfrac{x^{t+1}}{t+1}\right]_{x=0}^{1}=\dfrac{1}{t+1}\\ \dfrac{-1}{\left(t+1\right)^{2}}=\dfrac{\mathrm{d}}{\mathrm{d}t}\dfrac{1}{t+1}=\dfrac{\mathrm{d}}{\mathrm{d}t}\Im\left(t\right)=\Im^{\prime}\left(t\right)= & \dfrac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}x^{t}\mathrm{d}x=\int_{0}^{1}\dfrac{\partial x^{t}}{\partial t}\mathrm{d}x=\int_{0}^{1}x^{t}\ln x\mathrm{d}x\\ -1=\left[\dfrac{-1}{\left(t+1\right)^{2}}\right]_{t=0}=\Im^{\prime}\left(0\right)= & \int_{0}^{1}x^{0}\ln x\mathrm{d}x=\int_{0}^{1}1\ln x\mathrm{d}x=\int_{0}^{1}\ln x\mathrm{d}x\\ \int_{0}^{1}\ln x\mathrm{d}x= & -1 \end{aligned}

https://zhuanlan.zhihu.com/p/687355703

https://www.zhihu.com/question/646881575/answer/3417318090

TaylorCatAlice: Feynman method of integration and residue theorem

https://www.bilibili.com/video/BV1Lj411L79X

MatheManiac: complex integral

https://www.youtube.com/watch?v=EyBDtUtyshk&list=PLDcSwjT2BF_UDdkQ3KQjX5SRQ2DLLwv0R&index=11

https://www.youtube.com/watch?v=EyBDtUtyshk