Limits that occur in integrals during Feynmann integration

Question

When we use Feynman Integration, how do we decide the final constant of integration? For example, in this problem:

$$ f(a) = \int_{0}^{1} \frac{\arctan(ax)}{\sqrt{1-x^2}} dx$$

After differentiating and integrating, $$ f'(a) = \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}} $$ So we need to find out $$ f(1) = \int_{0}^{1} \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}} da$$

How were the limits of the definite integral decided?

Answer 1

This just uses the Fundamental Theorem of Calculus, which gives us that

$$f(1) - f(0) = \int_0^1 f'(a) \,da.$$

For $$f(a) = \int_0^1 \frac{\arctan(ax)}{\sqrt{1 - x^2}} dx,$$ we have $f(0) = \int_0^1 0 \,dx = 0,$ and substituting gives the last display equation in the question.

Answer 2

You pick upper bound such that it matches the original integrand, and the lower bound such that it makes the original integrand 0.