Tricky Definite Integral?

If u>0 let J(u) = ∫ [x=0,∞] ( exp(−x)−exp(−ux) )/x dx (convergent if u>0)

dJ/du = ∫ [x=0,∞] x.exp(−ux) )/x dx = ∫ [x=0,∞] exp(−ux) dx = 1/u

∴ J(u) = logₑ(u) + C

Because J(1)=0 so then C=0 which gives J(2)=logₑ(2)

Its been a together as yet i think of you have separate those: eval ? x dx from -a million to a million and ? a million/x dx from a million to 5 x²/2 (from -a million to a million) and ln x (from a million to 5) a million/2 - a million/2 = 0 ln 5 - ln a million = a million.609 0 + a million.609 = a million.609 This demands a non-end functionality at x = a million, which it is not. ie the two applications have distinctive y values at x = a million. i'm undecided the place to circulate from right here, yet i'm hoping i've got helped.

∫ [e^(-x) - e^(-2x)] / x dx

Separate to two parts.

1.

∫ e^(-x) / x dx

= ∫ e^(-x) dlnx ------- dlnx = 1/x dx

= lnxe^(-x) - ∫ d(e^(-x)lnx ----- ∫ udv = uv - ∫ duv

= lnxe^(-x) + ∫ lnxe^(-x)dx ------- d(e^(-x) = - e^(-x)dx

2. Similarly

∫ e^(-2x) / x dx

= ∫ e^(-2x) dlnx

= lnxe^(-2x) - ∫ de^(-2x) lnx

= lnxe^(-2x) + 2∫ lnxe^(-2x)dx

For 2∫ lnxe^(-2x)dx,

Let u = 2x. du = 2dx, as x→∞, u→∞

2∫ lnxe^(-2x)dx

= 2∫ ln(u/2)e^(-u) du/2

= ∫ (lnu - ln2)e^(-u) du

= ∫ lnu e^(-u) du - ∫ ln2e^(-u) du

= ∫ lnu e^(-u) du + ln2e^(-u)

Thus,

∫ e^(-2x) / x dx = lnxe^(-2x) + ∫ lnu e^(-u) du + ln2e^(-u) -------- (1)

From 1., we know

∫ e^(-x) / x dx = lnxe^(-x) + ∫ lnxe^(-x)dx ------- (2)

(2) - (1)

∫ e^(-x) / x dx - ∫ e^(-2x) / x dx

= lnxe^(-x) - lnxe^(-2x) + ∫ lnxe^(-x)dx - ∫ lnu e^(-u) du - ln2e^(-u)

Now do the definite integral, x from 0 to ∞

All terms, but ln2, cancel out.

Thus, THE ANSWER IS ln2

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