Our "Mathematical Analysis" tutorial contains two quite interesting definite integrals. After testing, they can be solved using the logic of Feynman’s integration method.
\begin{aligned}\int_0^1 \frac{\ln(1+x)}{1+x^2}dx \\ \int_0^{\pi} \frac{x \sin x}{1+\cos^2 x}dx\end{aligned}
No.1
First, let’s consider the first one. Actually, the idea for this problem is quite obvious because the derivative of a logarithmic function becomes a rational fraction, and there is a unified method for handling integrals of this form. Thus:
\begin{aligned}F(a)&=\int_0^1 \frac{\ln(1+ax)}{1+x^2}dx \\ F'(a)&=\int_0^1 \frac{x}{(1+ax)(1+x^2)}dx=\int_0^1 \frac{1}{1+a^2}\left(\frac{-a}{1+ax}+\frac{x+a}{1+x^2}\right)dx \\ &=-\frac{\ln(1+a)}{1+a^2}+\frac{\ln 2}{2}\left(\frac{1}{1+a^2}\right)+\frac{\pi}{4} \frac{a}{1+a^2}\end{aligned}
Next, we integrate with respect to a. It is easy to see that F(0)=0, therefore: F(a)=\int_0^a -\frac{\ln(1+t)}{1+t^2}dt + \frac{\ln 2}{2} \arctan(a)+\frac{\pi}{8}\ln(1+a)
So \begin{aligned}F(1)&=\int_0^1 -\frac{\ln(1+t)}{1+t^2}dt + \frac{\pi}{4}\ln 2 \\ &=-F(1)+ \frac{\pi}{4}\ln 2\end{aligned}
Therefore F(1)=\frac{\pi}{8}\ln 2.
No.2
In the second integral, the author (who is our Mathematical Analysis teacher) provided a clever solution. This solution makes high use of the symmetry of trigonometric functions and is worth learning and imitating. The original proof will be given at the end. Here, we first explore the process of solving it using Feynman’s integration method.
We have: \int_0^{\pi} \frac{x \sin x}{1+\cos^2 x}dx=-\int_0^{\pi} \frac{x}{1+\cos^2 x}d(\cos x)
It would be great if that single term x, after differentiation, also became a function of \cos x; this would be equivalent to an ordinary polynomial integration. That’s easy to handle—let x=\arccos(a \cos x), i.e.,
F(a)=-\int_0^{\pi} \frac{\arccos(a \cos x)}{1+\cos^2 x}d(\cos x)
Wait, that’s not right. x traverses [0,\pi], but the range of the function \arccos(x) is [-\frac{\pi}{2},\frac{\pi}{2}], which clearly does not correspond. So, we must make a small modification: F(a)=-\int_0^{\frac{\pi}{2}} \frac{\arccos(a \cos x)}{1+\cos^2 x}d(\cos x)-\int_{\frac{\pi}{2}}^{\pi} \frac{\arccos(a \cos x)+\frac{\pi}{2}}{1+\cos^2 x}d(\cos x)
That is: F(a)=-\int_0^{\pi} \frac{\arccos(a \cos x)}{1+\cos^2 x}d(\cos x)- \frac{\pi}{2}\int_{\frac{\pi}{2}}^{\pi}\frac{1}{1+\cos^2 x}d(\cos x)
The second part on the right side of the equation is easy to integrate, resulting in -\frac{\pi^2}{4}. The key is the integral on the left: f(a)=-\int_0^{\pi} \frac{\arccos(a \cos x)}{1+\cos^2 x}d(\cos x)
After differentiation, it becomes: \begin{aligned}f'(a)&=-\int_0^{\pi} \frac{\cos x}{\sqrt{1-a^2 \cos^2 x}(1+\cos^2 x)}d(\cos x) \\ &=-\frac{1}{2} \int_0^{\pi} \frac{1}{\sqrt{1-a^2 \cos^2 x}(1+\cos^2 x)}d(\cos^2 x)\end{aligned}
Let t=\cos^2 x, then it becomes: f'(a)=-\frac{1}{2}\int_1^1 \frac{1}{\sqrt{1-a^2 t}(1+t)}dt
Since the upper and lower limits of integration are equal, the integral value is clearly 0!!! Therefore, f(a) is a constant independent of a. Based on: f(0)=-\frac{\pi}{2}\int_0^{\pi} \frac{1}{1+\cos^2 x}d(\cos x)=\frac{\pi^2}{2}
We conclude that f(a)=\frac{\pi^2}{2}.
So F(1)=F(a)=f(a)-\frac{\pi^2}{4}=\frac{\pi^2}{4}.
Wonderful, isn’t it? Even where symmetry plays a role, Feynman’s integration method can still hold its ground!
Appendix: Solution from the book I=\int_0^{\pi} \frac{x \sin x}{1+\cos^2 x}dx Let x=\pi-t, then \begin{aligned}I&=\int_{\pi}^0 \frac{(\pi-t) \sin (\pi-t) }{1+\cos^2 (\pi-t)}d(\pi-t) \\ &=\int_0^{\pi} \frac{\pi \sin t }{1+\cos^2 t}dt-\int_0^{\pi} \frac{t \sin t }{1+\cos^2 t}dt \\ &=\frac{\pi^2}{2}-I\end{aligned}
So I=\frac{\pi^2}{4}.
When reposting, please include the address of this article: https://kexue.fm/archives/1944
For more detailed reposting matters, please refer to: Scientific Space FAQ