The Feynman Integral Trick — Differentiation Under the Integral Sign (4)

费曼积分法——积分符号内取微分(4) 苏剑林 · 2012-06-26 · Mathematics integral differential

Since I have some free time this morning, let me get this post written quickly. This afternoon the college entrance exam (gaokao) results will be announced, and after that I'll probably be busy again for a while. This should be the last post in the "Feynman integral trick" series. It's mainly about another worked example of the technique. What's different this time is that this is the first instance where BoJone independently solved a problem using the Feynman integral trick. All the previous examples were ones where the book provided hints, and BoJone worked them out with that guidance. So this problem holds a bit of special significance for me.

On the Mathematics Research and Development Forum, someone named Wayne once conjectured the following proposition:

$\int_0^{\infty}\frac{f(x,2m-1)-\sin x}{x^{2m+1}}dx$ where f(x, 2m-1) denotes the (2m-1)-th order Taylor expansion of sin x.
For example, when m=1,
$$\int_0^{\infty}\frac{x-\sin x}{x^3}dx$$
when m=2,
$$\int_0^{\infty}\frac{x-\frac{x^3}{6}-\sin x}{x^5}dx$$
With the help of some software, I found that the result is:
$\frac{\pi(-1)^{m-1}}{2(2m)!}$

Someone gave a proof using the Laplace transform, but I only have a conceptual understanding of the Laplace transform and don't know the specific method. So I used the Feynman integral trick to generalize the problem. Suppose we have a function y=f(x), and we want to compute the integral

$$F(t)=\int_a^b \frac{f(0)+f'(0)(tx)+f''(0)\frac{(tx)^2}{2}+...+f^{(n)}(0)\frac{(tx)^n}{n!}-f(tx)}{x^{n+1}} dx$$

We have

$$\begin{aligned}\frac{d F(t)}{dt}=\int_a^b \frac{f'(0)+f''(0)(tx)+...+f^{(n)}(0)\frac{(tx)^{n-1}}{(n-1)!}-f'(tx)}{x^n} dx\end{aligned}$$

Differentiating n times in succession gives

$$\begin{aligned}\frac{d^n F(t)}{dt^n}=\int_a^b \frac{f^{(n)}(0)-f^{(n)}(tx)}{x} dx\end{aligned}$$

In some cases, we can differentiate n+1 times to get

$$\begin{aligned}\frac{d^{n+1} F(t)}{dt^{n+1}}=\int_a^b -f^{(n+1)}(tx) dx=-f^{(n)} (tx)|_a^b\end{aligned}$$

Then we integrate with respect to the variable t either n times or n+1 times, a process that introduces several integration constants. Since we know the original integral equals 0 when t=0, we can determine that every one of these integration constants is 0.

For Wayne's problem, f(x) = sin(x), n = 2m, $f^{(2m)}(x)=(-1)^m sin x$

$$\begin{aligned}\frac{d^{2m} F(t)}{dt^{2m}}=\int_0^{\infty} \frac{-(-1)^m \sin (tx)}{x} dx\end{aligned}$$

(at this point we can no longer differentiate further — doing so would give $-(-1)^m sin (tx)|_0^{\infty}$, which doesn't make sense here)

Based on $\int_0^{\infty} \frac{sinx}{x}dx=\frac{\pi}{2}$

we have

$$\frac{d^{2m} F(t)}{dt^{2m}}=-(-1)^m\frac{\pi}{2}=(-1)^{m-1} \frac{\pi}{2}$$

After integrating 2m times: $F(t)=\frac{(-1)^{m-1} \pi \cdot t^{2m}}{2(2m)!}$

Just set t=1.

And so, an interesting definite integral has been elegantly resolved by the Feynman integral trick!

Appendix: Proof via the Laplace Transform

Laplace transform proof

Conclusion

Most standard calculus textbooks nowadays no longer teach "differentiation under the integral sign" as a tool. To deal with more complicated integrals, people generally reach for tools from complex analysis (namely the "contour integration" mentioned earlier), or else just plug things into a computer and be done with it. Still, BoJone wanted to introduce readers to this rather delightful technique, so that people know it exists, and know something of its history too. After all, the "boy wonder" Feynman used it to crack quite a few tough problems — how could we resist giving it a try ourselves?