Integral identity?

Question

Here's one inspired by a question of Eulercrosser:

Evaluate
int [exp(-ax)-exp(-bx)]/x dx
where a,b>0 and the integral goes from 0 to infinity.

It isn't an answer unless it is justified.

fatal_flad_death has the correct answer. Now show why it is correct.

Answer 1

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)

Answer 2

hi, is or become this considered one of YOUR homework assignments? you're a competent one for symantics of the questions too, and as you probably did no longer ask to instruct it, yet a thank you to instruct it, right this is an answer, and doubtless a sprint for others. word that ? (x^a -a million)/log(x) dx = ?(x^a)/log(x) dx - ?(a million/log(x)) dx permit y=log(x) ? x = e^y , dx = e^y dy then ? = ? (e^(ay))/y * e^y dy - ?(a million/log(x)) dx = ? (e^[(a+a million)y])/y dy - ?(a million/log(x)) dx = Ei[(a+a million)y] - Li[x] =Ei[(a+a million)log(x)] - Li[x] use Li[z]=Ei[log(z)], and circulate from there... Eulercrosser, now circulate verify my answer to: Why is the diff of two numbers made by potential of a diverse mixtures of the comparable digits, equivalent to a diverse of 9?

Answer 3

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

Answer 4

Is it (b-a)