For - min(alpha,beta)<a ≤min(alpha,beta), expand cos(ax)

For $-\min(\alpha,\beta)

Step 1 · Expand and interchange

$$ I = \int_0^{\infty}\frac{e^{-\alpha x}-e^{-\beta x}}x\sum_{n=0}^{\infty}\frac{(-1)^na^{2n}x^{2n}}{(2n)!}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^na^{2n}}{(2n)!}\int_0^{\infty}(e^{-\alpha x}-e^{-\beta x})x^{2n-1}\,dx. $$ Interchange is justified since $|a|\le\min(\alpha,\beta)$ keeps the exponential decay dominant, so the series of integrals converges uniformly (Weierstrass M-test with the same bound used in Problem 281's remainder estimate).

Step 2 · Evaluate each integral

For $n=0$ (a case the naive Gamma-function formula does not directly cover, since $x^{-1}$ alone is non-integrable at $0$ and $\infty$): the combination is the classical Frullani integral, $$ \int_0^{\infty}\frac{e^{-\alpha x}-e^{-\beta x}}x\,dx = \ln\frac\beta\alpha. $$ For $n\ge1$, using $\int_0^{\infty}e^{-px}x^{2n-1}dx=(2n-1)!/p^{2n}$: $$ \int_0^{\infty}(e^{-\alpha x}-e^{-\beta x})x^{2n-1}\,dx = (2n-1)!\left(\frac1{\alpha^{2n}}-\frac1{\beta^{2n}}\right). $$

Step 3 · Resum the series

$$ I = \ln\frac\beta\alpha + \sum_{n=1}^{\infty}\frac{(-1)^na^{2n}}{2n}\left(\frac1{\alpha^{2n}}-\frac1{\beta^{2n}}\right) = \ln\frac\beta\alpha - \frac12\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n\left(\frac a\alpha\right)^{2n} + \frac12\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n\left(\frac a\beta\right)^{2n}. $$ Recognizing each sum as $\ln(1+t)$ with $t=(a/\alpha)^2$ and $(a/\beta)^2$: $$ I = \ln\frac\beta\alpha - \frac12\ln\left(1+\frac{a^2}{\alpha^2}\right) + \frac12\ln\left(1+\frac{a^2}{\beta^2}\right) = \frac12\ln\frac{\beta^2+a^2}{\alpha^2+a^2}. $$ (The $\ln\beta-\ln\alpha$ terms combine exactly with the two half-logarithms to leave this clean form — verified by expanding all logarithms explicitly.)

Summary Table

Additional Notes

• At $a=0$: $I=\frac12\ln(\beta^2/\alpha^2)=\ln(\beta/\alpha)$, recovering the ordinary Frullani integral exactly — a built-in consistency check on the general formula.
• Differentiating under the integral sign (the Feynman-trick route) gives the same answer more quickly: $\frac{dI}{da}=-\int_0^\infty(e^{-\alpha x}-e^{-\beta x})\sin(ax)\,dx=\frac a{\beta^2+a^2}-\frac a{\alpha^2+a^2}$, which integrates to the same closed form — a useful cross-check on the more laborious series computation.
• The key subtlety making the series approach work is recognizing that the $n=0$ term is not literally "$\int x^{-1}(\cdots)dx$ diverges" but rather the finite, well-defined Frullani integral — the two exponentials' difference regularizes the apparent singularity at $x=0$.