Playing “Feynman’s Trick” on Indefinite Integrals – Tongue in Cheek

“Differentiation under the integral sign”, a.k.a., “Feynman’s trick” is a clever application of Leibniz’s rule (LR-1):

Let f(x, \beta) be continuous and have a continuous derivative \frac{\partial}{\partial \beta} in a domain of x\beta-plane that includes the rectangle a \le x \le b, \beta_1 \le \beta \le \beta_2,

\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx =\int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x, \beta)\;dx.

“Feynman’s trick” is known to be an effective technique for evaluating difficult definite integral such as \int\limits_{0}^{1}\frac{x-1}{\log(x)}\;dx.

Is Feynman’s “trick” applicable to indefinite integrals too?

In other words, is it also true that

\frac{\partial}{\partial \beta}\int f(x, \beta)\;dx + C = \int \frac{\partial}{\partial \beta}f(x, \beta)\;dx?\quad\quad\quad(\star)


Let’s apply(\star) to indefinite integral \int \log(x)\;dx:

\frac{\partial}{\partial \beta}\int x^{\beta}\;dx+C = \int \frac{\partial}{\partial \beta}x^{\beta}\;dx = \int x^{\beta}\log(x)\;dx;

i.e.,

\int x^{\beta}\log(x)\;dx =\frac{\partial}{\partial \beta}\int x^{\beta}\;dx+C.\quad\quad\quad(1)

Since \int x^{\beta}\; dx = \frac{x^{\beta+1}}{\beta+1} + C_1, the right-hand side of (1) is

\frac{\partial}{\partial \beta}\left(\frac{x^{\beta+1}}{\beta+1}+C_1\right) + C= \frac{x^{\beta+1}\log(x)\cdot (\beta+1) - x^{\beta+1}}{(\beta+1)^2}+C.

It means

\int x^{\beta}\log(x)\;dx = \frac{x^{\beta+1}\log(x)\cdot (\beta+1) - x^{\beta+1}}{(\beta+1)^2}+C.

For \beta = 0, we have

\int \log(x)\;dx = x\log(x)-x+C.

It checks out:

\frac{d}{dx}(x\log(x)-x+C) = \log(x)+x \cdot \frac{1}{x}-1 = \log(x).

Let’s also evaluate \int x e^{x}\;dx:

By (\star),

\frac{\partial}{\partial \beta}\int e^{\beta x}\;dx + C  = \int\frac{\partial}{\partial \beta} e^{\beta x}\;dx=\int x e^{\beta x}\;dx.

That is,

\int x e^{\beta x}\;dx = \frac{\partial}{\partial \beta}\int e^{\beta x}\; dx+C= \frac{\partial}{\partial \beta}\left(\frac{1}{\beta}e^{\beta x} + C_1\right)+C=\frac{e^{\beta x}\beta x - e^{\beta x}}{\beta^2}+C.\quad\quad(2)

Let \beta = 1, (2) yields

\int x e^{x} \;dx= e^x (x-1)+C.

It checks out too:

\frac{d}{dx} (e^x (x-1)+C) = e^x(x-1) + e^x = x e^x.

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Now that we have gained confidence in the validity of (\star), let’s prove it.

Given

G_1(x) = \int g(x)\;dx, G_2(x) = \int\limits_{a}^{x} g(x)\; dx

where g(x) is a function of x, we have,

(G_1(x)-G_2(x))' = (\int g(x)\;dx)' -  (\int\limits_{a}^{x}  g(x) \;dx)'= g(x)-g(x) = 0.

It means

G_1(x)-G_2(x)=C\implies \int g(x)\;dx= \int\limits_{a}^{x}g(x)\;dx + C.

When x=b,

\int g(x)\;dx = \int\limits_{a}^{b}g(x)\;dx+C;

i.e., for f(x,\beta), a function of both x and \beta,

\int f(x,\beta)\;dx = \int\limits_{a}^{b} f(x, \beta)\;dx+C\quad\quad\quad(3)

and,

\int \frac{\partial}{\partial t} f(x,\beta)\;dx = \int\limits_{a}^{b} \frac{\partial}{\partial \beta}f(x, \beta)\;dx+C.\quad\quad\quad(4)

It follows that

\frac{\partial}{\partial \beta}\int f(x,\beta)\;dx \overset{(3)}{=} \frac{\partial}{\partial \beta}\left(\int\limits_{a}^{b}f(x,\beta)\;dx+C\right)

=\frac{\partial}{\partial \beta}\int\limits_{a}^{b}f(x,\beta)\;dx

\overset{\textbf{LR-1}}{=} \int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x,\beta)\;dt

\overset{(4)}{=} \int\frac{\partial }{\partial \beta}f(x,\beta)\;dx -C.

And so,

\frac{\partial}{\partial \beta}\int f(x,\beta)\;dx +C= \int \frac{\partial}{\partial \beta} f(x,\beta)\;dx.


Exercise-1 Evaluate \int x^2 e^x\;dx.

hint: \frac{\partial}{\partial \beta}\int x e^{\beta x}\;dx = \int x^2 e^{\beta x}\; dx.

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