# Translating integrals by scalar premultiplication

Say I have the integral $$I=\int_a^b e^{f(x)} \text{d}x,$$ if I premultiply $I$ by $e^{\delta}$, I can translate the argument of the exponential inside the integral, e.g.

$$e^\delta I = \int_a^b e^{f(x)+\delta} \text{d}x.$$

What's important/cool here, is that I can work with just the value of $I$, for example $I$ could be found from numerical integration of some complicated $f(x)$. So I've done a different integral, without having to integrate again.

Is there a way to generalise this beyond just the exponential? For example I'm working with the hyperbolic tangeant, $$J = \int^b_a \tanh{\left( f(x)\right)}\text{d}x.$$

I was hoping, that by expressing $\tanh(y)=\frac{e^{2y}-1}{e^{2y}+1}$, and then expanding it and working with each individual part, I'd be able to find a way to achieve a similar effect, but I'm struggling with dealing with the denominator.

The question then, is: Is there a way to take $J\rightarrow \int^b_a \tanh{\left( f(x) +\delta \right)}\text{d}x$, without needing to re-do the integral $J$?

• $$\tanh(f(x) + \delta) = \frac{\tanh f(x) + \tanh \delta}{1 + \tanh f(x) \tanh \delta}$$ Im not sure you can do it – Ziad Fakhoury Aug 21 '17 at 14:46