Question #23877

Question

Question #23877

1 Answer

Impact of this question

2304 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-20T15:47:13

$I = \int_{0}^{1} \frac{x^{b} - x^{a}}{ln (x)} d x = ln (\frac{b + 1}{a + 1})$ $I = int_0^1 (x^b-x^a)/(ln(x))"d"x = ln((b+1)/(a+1))$ .

Explanation:

$I = \int_{0}^{1} \frac{x^{b} - x^{a}}{ln (x)} d x$ $I = int_0^1 (x^b-x^a)/(ln(x))"d"x$

Make the substitution $u = - ln (x)$ $u=-ln(x)$ . Then $\frac{d u}{d x} = - \frac{1}{x}$ $("d"u)/("d"x)=-1/x$ , $d x = - x d u$ $"d"x=-x"d"u$ . As $x \to 0$ $x -> 0$ , $u \to + \infty$ $u -> + \infty$ . And at $x = 1$ $x=1$ , $u = 0$ $u=0$ .

Then the integral is transformed to,

$I = \int_{\infty}^{0} \frac{x^{b} - x^{a}}{ln (x)} (- x) d u$ $I = int_(\infty)^(0) (x^b-x^a)/(ln(x)) (-x) "d"u$ .

As $u = - ln (x)$ $u=-ln(x)$ , $x = e^{- u}$ $x=e^(-u)$ .

Then,

$I = \int_{\infty}^{0} \frac{e^{- b u} - e^{- a u}}{- u} (- e^{- u}) d u$ $I= int_(\infty)^0 (e^(-bu)-e^(-au))/(-u) (-e^(-u)) "d"u$ ,
$I = \int_{\infty}^{0} \frac{e^{- (b + 1) u} - e^{- (a + 1) u}}{u} d u$ $I = int_(\infty)^(0) (e^(-(b+1)u)-e^(-(a+1)u))/(u) "d"u$ .

Integrand can be negated and the limits of integration reversed.

$I = \int_{0}^{\infty} \frac{e^{- (a + 1) u} - e^{- (b + 1) u}}{u} d u$ $I = int_(0)^(\infty) (e^(-(a+1)u)-e^(-(b+1)u))/(u) "d"u$ .

This is now, surprisingly, in a standard form. It is called Frullani's integral. It states that

$\int_{0}^{\infty} \frac{f (a x) - f (b x)}{x} d x = (f (\infty) - f (0)) ln (\frac{a}{b})$ $int_(0)^(infty) (f(ax)-f(bx))/(x) "d"x = (f(infty)-f(0))ln(a/b)$

if $f$ $f$ is such that $f (\infty)$ $f(\infty)$ and $f (0)$ $f(0)$ exists. If we choose $f (x) = e^{- x}$ $f(x)=e^(-x)$ , this is transformed into $I$ $I$ , with $f (\infty) = 0$ $f(\infty)=0$ and $f (0) = 1$ $f(0)=1$ .

We conclude,

$\int_{0}^{\infty} \frac{e^{- (a + 1) u} - e^{- (b + 1) u}}{u} d u = - ln (\frac{a + 1}{b + 1})$ $int_(0)^(\infty) (e^(-(a+1)u)-e^(-(b+1)u))/(u) "d"u = -ln((a+1)/(b+1))$ ,
$\int_{0}^{\infty} \frac{e^{- (a + 1) u} - e^{- (b + 1) u}}{u} d u = ln (\frac{b + 1}{a + 1})$ $int_(0)^(\infty) (e^(-(a+1)u)-e^(-(b+1)u))/(u) "d"u = ln((b+1)/(a+1))$ ,
$I = ln (\frac{b + 1}{a + 1})$ $I = ln((b+1)/(a+1))$ .

Science

Math

Humanities

... and beyond

Question #23877

1 Answer

Explanation:

Related questions

Impact of this question