Integral identity?

Yes.

Switching a,b only changes the sign, so let's suppose b < a without loss of generality. Write the integrand:

exp(-ax)[1-exp(a-b)x]/x

= -exp(-ax)[(a-b) + (a-b)^2x/2! + (a-b)^3x^2/3! + ...]

Since \int exp(-ax)x^n dx= n!/a^(n+1) , the integral is:

-[(a-b)/a + (a-b)^2/2a^2 + (a-b)^3/3a^3 + ...]

Since 0 < (a-b)/a <= 1 this the the series for the logarithm:

ln(1-(a-b)/a) = ln(b) - ln(a)

*************

Another way to see it, more informally:

W(a) = \int (epsilon-->oo) exp(-ax)/x dx

dW(a)/da = -exp(-a epsilon)/a ~ -1/a + epsilon +...

==>

W(a) + Constant - ln(a)

==>

W(a) - W(b) = ln(b) - ln(a)

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I believe that it is ln(b)-ln(a)

Is it (b-a)