Feynmann's technique for integration

Feynmann's technique is a technique for evaluating certain difficult definite integrals. Note definite integral here, it doesn't do anything to help find antiderivatives. In fact, Feynmann's technique is especially helpful with finding definite integrals that have no elementary antiderivative.

The formula for Feynmann's technique is surprisingly simple:

$$ \frac{d}{dt} \int f(x, t) dx = \int \frac{\partial f}{\partial t} f(x, t) dx $$

And the method is too - just 4 steps:

Consider:

$$ \int \limits_0^1 \ln(x) dx $$

To use Feynmann's trick, we need to add a second variable $t$ to a convenient place inside our function, making a new function $I(t)$. Here there's really only one place to put it, so we define $I(t)$ as:

$$ I(t) = \int \limits_0^1 \ln(xt) dx $$

Now notice $I(1)$ is equal to our original integral:

$$ I(1) = \int \limits_0^1 \ln(x \cdot 1)dx = \int \limits_0^1 \ln(x) dx $$

So the key to evaluating our integral is to find out what $I(1)$ is. But surely this isn't easy, you say? Didn't we just make our original integral more complicated?

At first glance, yes. But notice that we can take the derivative of $I$, to get:

$$ I'(t) = \frac{d}{dt} \int \limits_0^1 \ln(tx) dx = \int \limits_0^1 \frac{\partial}{\partial t}\ln (tx) dx $$

$$ I'(t) = \int \limits_0^1 \frac{1}{t} dx = \frac{1}{t} $$

Now, we can evaluate $I(t)$ by taking the integral of $I'(t)$:

$$ I(t) = \int I'(t) dt = \ln(t) + C $$

We can find out $C$ by finding an appropriate value of $t$ that will yield a constant for $I(t)$. This is really the biggest limitation of Feynmann's integral trick - you have to have functions in a form where substituting in a value for $t$ cancels out an integral and leaves you with a number. In our case, we'll skip ahead and just note that $C = -1$. So:

$$ I(1) = \ln(1) + C = 0 + (-1) = -1 $$

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