This term can thus be rewritten using derivatives of the exponential function. The goal is to determine the factors xk to replace from the equation. The form then looks more complicated, but proves to be extremely useful for subsequently simplifying the integral. Here the practical property is crucial that the derivative of the exponential function is again the exponential function:
[ f(x) = lim_epsilon to 0sum_k=0^infty a_k x^k e^epsilon x = lim_epsilon to 0sum_k=0^infty a_k left(fracdd epsilonright)^k e^epsilon x]
With each derivation of the exponential function with respect to ε one obtains according to the chain rule a x as a factor. You’re almost there. Because as it turns out, you can get the infinite series back into the original function f rewrite. However, the argument changes: the power series no longer depends on x but from the derivative with respect to ε :
[ f(x) = lim_epsilon to 0sum_k=0^infty a_k left(fracdd epsilonright)^k e^epsilon x = lim_epsilon to 0 f left(fracdd epsilonright) e^epsilon x ]
You’ve already reached your goal. By substituting this function into an integral, one can make the simplifications to get rid of the integration entirely. Then f no longer depends on the integration variable x and can thus be extracted from the integral. You only have to integrate over an exponential function – the simplest argument imaginable for an integral:
[ int_a^b f(x) dx = lim_epsilon to 0 f left(fracdd epsilonright) int_a^b e^epsilon x dx = lim_epsilon to 0 f left(fracdd epsilonright) frace^epsilon b-e^epsilon aepsilon ]
In this way, the complicated part of integrating can be replaced by a derivative. The difficulty of the task now is to find out what f(d/dε). That helps here f can be rewritten as a power series: (sum_k=0^infty a_k left(fracdd epsilonright)^k ). That is, one gets terms of higher derivatives applied to the expression (frace^epsilon b-e^epsilon aepsilon ).
For example, you can ask yourself what will happen if f is an exponential function. In other words: What is the result of ed/dε applied to any differentiable function G(ε)? As it turns out (see box below), the result is nothing more than a shift in function G:ed/dεG(ε) = G(ε+1).
Since many functions (e.g. the sine, the cosine or hyperbolic functions) can be expressed by exponential functions, their integrals can be calculated very quickly using the new method – even if many of them are linked together. Where else the partial integration is used again and again, you can now cleverly calculate the derivatives instead.