Calculus – Integration – The Feynman method

The Feynman method is based on the following theorem:

Theorem: If \(f(x,t)\) and \(\dfrac{\partial}{\partial t}f(x,t)\) are both continuous for \(x\in[a,b]\), then we have:

\[\frac{d}{dt}\int_a^bf(x,t)\,dx=\int_a^b\frac{\partial}{\partial t}f(x,t)\,dx.\]

For the notation \(\dfrac{\partial}{\partial t}\) see: partial derivatives. The method is based on finding an integrand \(f(x,t)\) with a parameter \(t\) such that the derivative with respect to that parameter \(t\) leads to an integrand that is easier to integrate. Some examples:

1) Consider the integral \(\displaystyle\int_0^1\frac{x-1}{\ln(x)}\,dx\). Since \(\displaystyle\lim\limits_{x\downarrow0}\frac{x-1}{\ln(x)}=0\) and \(\displaystyle\lim\limits_{x\uparrow1}\frac{x-1}{\ln(x)}=1\) (see: l'Hospital's rule) the integrand is continuous on \([0,1]\) and the integral exists. Now consider the integral \(I(t)=\displaystyle\int_0^1\frac{x^t-1}{\ln(x)}\,dx\) for \(t>0\). Then we have:

\[I'(t)=\int_0^1\frac{x^t\ln(x)}{\ln(x)}\,dx=\int_0^1x^t\,dx=\frac{1}{t+1}x^{t+1}\bigg|_0^1=\frac{1}{t+1},\quad t>0.\]

This implies that \(I(t)=\ln(t+1)+C\) and since \(I(0)=0\) we conclude that \(I(t)=\ln(t+1)\) for \(t>0\).

Hence: \(\displaystyle\int_0^1\frac{x-1}{\ln(x)}\,dx=\ln(2)\). Similarly: \(\displaystyle\int_0^1\frac{x^2-1}{\ln(x)}\,dx=\ln(3)\) and \(\displaystyle\int_0^1\frac{\sqrt{x}-1}{\ln(x)}\,dx=\ln(\tfrac{3}{2})\).

2) Consider the integral \(\displaystyle\int_0^1\frac{\sin(\ln(x))}{\ln(x)}\,dx\). Let \(I(t)=\displaystyle\int_0^1\frac{\sin(t\ln(x))}{\ln(x)}\,dx\), then we have: \(I(0)-0\) and using the substitution \(\ln(x)=y\) or equivalently \(x=e^y\)

\[I'(t)=\int_0^1\cos(t\ln(x))\,dx=\int_{-\infty}^0\cos(ty)e^y\,dy=\int_0^{\infty}e^{-y}\cos(ty)\,dy.\]

Now we have using integration by parts

\begin{align*} \int_0^{\infty}e^{-y}\cos(ty)\,dy&=-e^{-y}\cos(ty)\bigg|_0^{\infty}-t\int_0^{\infty}e^{-y}\sin(ty)\,dy\\[2.5mm] &=1+te^{-y}\sin(ty)\bigg|_0^{\infty}-t^2\int_0^{\infty}e^{-y}\cos(ty)\,dy=1-t^2\int_0^{\infty}e^{-y}\cos(ty)\,dy. \end{align*}

Hence: \(I'(t)=\displaystyle\int_0^{\infty}e^{-y}\cos(ty)\,dy=\frac{1}{1+t^2}\). Since \(I(0)=0\) this implies that \(I(t)=\arctan(t)\).

Hence: \(\displaystyle\int_0^1\frac{\sin(\ln(x))}{\ln(x)}\,dx=\arctan(1)=\tfrac{1}{4}\pi\).

3) Consider the integral \(I(t)=\displaystyle\int_0^{\infty}\frac{e^{-\alpha x}\sin(xt)}{x}\,dx\) for \(\alpha>0\) and \(t>0\). Since \(\displaystyle\lim\limits_{x\downarrow0}\frac{\sin(xt)}{x}=t\), the integral is convergent. Then we have: \(I(0)=0\) and using integration by parts

\[I'(t)=\int_0^{\infty}e^{-\alpha x}\cos(xt)\,dx=\left[\frac{t}{\alpha^2+t^2}e^{-\alpha x}\sin(xt)-\frac{\alpha}{\alpha^2+t^2}e^{-\alpha x}\cos(xt)\right]_0^{\infty}=\frac{\alpha}{\alpha^2+t^2}.\]

This implies that \(I(t)=\arctan\left(\frac{t}{\alpha}\right)\). For instance, this implies that \(\displaystyle\int_0^{\infty}\frac{e^{-\alpha x}\sin(\alpha x)}{x}\,dx=\arctan(1)=\tfrac{1}{4}\pi\) for all \(\alpha>0\). Furthermore, we have: \(\displaystyle\int_0^{\infty}\frac{e^{-x\sqrt{3}}\sin(x)}{x}\,dx=\arctan\left(\tfrac{1}{\sqrt{3}}\right)=\tfrac{1}{6}\pi\) and \(\displaystyle\int_0^{\infty}\frac{e^{-x}\sin(x\sqrt{3})}{x}\,dx=\arctan(\sqrt{3})=\tfrac{1}{3}\pi\).

The Dirichlet integral

Theorem: \(\displaystyle\int_0^{\infty}\frac{\sin(x)}{x}\,dx=\frac{1}{2}\pi\).

Proof: Since \(\displaystyle\lim\limits_{x\downarrow0}\frac{\sin(x)}{x}=1\) it is clear that the integrand is continuous and the integral convergent. Now let: \(I(t)=\displaystyle\int_0^{\infty}e^{-xt}\frac{\sin(x)}{x}\,dx\). Then we have:

\begin{align*} I'(t)&=-\int_0^{\infty}e^{-xt}\sin(x)\,dx=\int_0^{\infty}e^{-xt}\,d\cos(x)=e^{-xt}\cos(x)\bigg|_0^{\infty}-\int_0^{\infty}\cos(x)\,de^{-xt}\\[2.5mm] &=-1+t\int_0^{\infty}e^{-xt}\cos(x)\,dx=-1+t\int_0^{\infty}e^{-xt}\,d\sin(x)\\[2.5mm] &=-1+te^{-xt}\sin(x)\bigg|_0^{\infty}-t\int_0^{\infty}\sin(x)\,de^{-xt}=-1+t^2\int_0^{\infty}e^{-xt}\sin(x)\,dt. \end{align*}

This implies that \(\displaystyle\int_0^{\infty}e^{-xt}\sin(x)\,dx=\frac{1}{1+t^2}\) and therefore: \(I'(t)=-\displaystyle\frac{1}{1+t^2}\). Hence: \(I(t)=-\arctan(t)+C\). Note that \(0=\displaystyle\lim\limits_{t\to\infty}I(t)=-\tfrac{1}{2}\pi+C\), which implies that \(C=\tfrac{1}{2}\pi\). Now we conclude that \(I(t)=-\arctan(t)+\frac{1}{2}\pi\). Hence: \(\displaystyle\int_0^{\infty}\frac{\sin(x)}{x}\,dx=I(0)=\tfrac{1}{2}\pi\).

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Last modified on April 20, 2024