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# Find all the values of k?

for which the definite integral:

∫ x^k ln(x) dx from 0 to 1 converges,

and evaluate the integral for those values of k.

### 3 Answers

- kbLv 75 years agoFavorite Answer
∫(x = 0 to 1) x^k ln x dx

= ∫(∞ to 0) (e^w)^k * w * (e^w dw), letting w = ln x <==> x = e^w

= -∫(w = 0 to ∞) we^((k+1)w) dw

= -∫(w = 0 to ∞) we^((k+1)w) dw.

(i) If k = -1, then this reduces to

-∫(w = 0 to ∞) w dw = -∞.

(ii) Otherwise, we can use integration by parts:

-∫(w = 0 to ∞) we^((k+1)w) dw

= -lim(t→∞) ∫(w = 0 to t) we^((k+1)w) dw

= -lim(t→∞) [we^((k+1)w)/(k+1) - e^((k+1)w)/(k+1)^2] {for w = 0 to t}

= (-1/(k+1)^2) * lim(t→∞) [(k+1)w - 1] e^((k+1)w) {for w = 0 to t}

= (-1/(k+1)^2) * lim(t→∞) [((k+1)t - 1) e^((k+1)t) + 1]

= (-1/(k+1)^2) * [1 + lim(t→∞) ((k+1)t - 1) e^((k+1)t)]

If k+1 > 0, then the limit is infinite.

If k+1 < 0, then we can apply L'Hopital's Rule:

(-1/(k+1)^2) * [1 + lim(t→∞) ((k+1)t - 1) e^((k+1)t)]

= (-1/(k+1)^2) * [1 + lim(t→∞) ((k+1)t - 1)/e^(-(k+1)t)]

= (-1/(k+1)^2) * [1 + lim(t→∞) (k+1)/(-(k+1) e^(-(k+1)t))]

= (-1/(k+1)^2) * [1 + lim(t→∞) -e^((k+1)t)]

= (-1/(k+1)^2) * (1 + 0), since k+1 < 0

= -1/(k+1)^2.

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In summary, the integral converges precisely when k+1 < 0 (or k < -1).

I hope this helps!

- Anonymous5 years ago
int_0^1 (x^k)*ln(x) dx

u = ln(x); du = 1/x dx; dv = x^k dx; v = x^(k + 1)/(k + 1)

[ln(x)*x^(k + 1)/(k + 1)] - int (x^k)/(k + 1) dx

[ln(x)*x^(k + 1)/(k + 1)] - [(x^(k + 1))/(k + 1)^2]

[x^(k + 1)/(k + 1)]*{[ln(x)] - [1/(k + 1)]}