I'm cross-posting from Physics site here because I don't know where this question fits better.
I'm trying to understand how the logarithm in eq. (3.25) from this paper appears from the equation above eq. (3.11).
Eq. (3.25) in paper is
$$ \int_0^1 d\lambda \int_0^1 dt\ f(\kappa t \lambda x) = -\int_0^1 d\lambda (\ln\lambda)f(\kappa \lambda x),\quad x = (x^0, \vec{x}), \quad \kappa, \lambda, t \in \mathbb{R}\tag1 $$
Equation above eq. (3.11) in paper is:
$$ \sum_{n = 0}^{\infty} \left[ 1 + \sum_{k = 1}^\infty\int_0^1 dt\ \left(\frac{-x^2}{4}\right)^k \frac{(\partial^2)^k t^n}{k!(k-1)!}\left(\frac{1-t}{t}\right)^{k-1} \right] \frac{(\kappa \lambda)^n}{n!}\bar{\psi}(0)\not{x} (xD)^n \psi(0) \tag2$$
According to the text, $\partial^2$ acts only over 4-vector $x$; and in the sum on $k$ only $k = 1$ contributes, so if you particularize for $k = 1$ the previous expression you are left with the polynomical integral
$$ \int_0^1 dt\ t^n = \frac{1}{n+1} \tag3 $$
So the operator in our eq. (2) is,
$$ \sum_{n = 0}^{\infty} \left[\frac{1}{n!} + \left(\frac{-x^2}{4}\right) \partial^2 \frac{1}{(n+1)!}\right] (\kappa \lambda)^n \bar{\psi}(0)\not{x} (xD)^n \psi(0) \tag4 $$
Function $f$ in our eq. (1) (eq. (3.25) in the paper) is precisely
$$ f(\kappa \lambda x) = \sum_{n = 0}^\infty \frac{(\kappa \lambda)^n}{n!}\bar{\psi}(0)\not{x} (xD)^n \psi(0) $$
I don't see how our eq. (4) can generate a logarithm that could lead you to eq. (3.25) in the paper, here eq. (1). To be more precise, the minus logarithm should appear in the the 2nd term of the sum on $n$ in eq. (4).
The relation in Equation (3.25) doesn’t require any particular properties of $f(x)$. You just need to change variables and swap the order of integration:
\begin{align*} \int_0^1\mathrm{d}\lambda \int_0^1\mathrm{d}t f(\kappa t \lambda x) &= \int_0^1\mathrm{d}t \int_0^t\frac{\mathrm{d}\lambda}{t}f(\kappa \lambda x),\\ &= \int_0^1\mathrm{d}\lambda \int_\lambda^1\frac{\mathrm{d}t }{t} f(\kappa \lambda x),\\ &= \int_0^1\mathrm{d}\lambda [\mathrm{ln}(1) -\mathrm{ln}(\lambda)] f(\kappa \lambda x),\\ &= -\int_0^1\mathrm{d}\lambda (\mathrm{ln}\lambda) f(\kappa \lambda x).\end{align*}