Evaluation of Integrals Using Gamma Function

Question

Question:

Evaluate the following integral using Gamma function.

Integrate x from 0-->1 dx/((-ln(x))^(1/2))

My approach:

I let -ln(x) = u ==> ln(x) = -u ==> x = e^(-u)

==> dx = -(e^(-u))

Limits of u: 0-infinty, which gives,

Integrate u from 0-->1 dx/((-ln(x))^(1/2)) = Integrate u from 0-->infinity { - du/((u^(1/2))*(e^(-u))) }

= -Gamma function[ (-1/2) + 1 ] = -(-1/2) * Gamma function [ (-1/2) ]

= (1/2) * (-2 * sqrt(pi) )

= -sqrt(pi)

Is this approach correct?

Thank you in advance!

Answer 1

You have done the right substitution, -ln(x) =u.

You wrote "dx = - e^(-u)"

Maybe you meant to say dx = -e^(-u) du?

Let us call our integral I. So you then wrote

"I = Integrate u from 0-->1 dx/((-ln(x))^(1/2)) = Integrate u from 0-->infinity { - du/((u^(1/2))*(e^(-u))) }"

I think there is an error here.

When x = 1, then -ln(x) = -ln(1) = 0 => u = 0, when x = 1.

When x = 0, -ln(x) approaches infinity => u = ∞, when x = 0.

Also note ∫(a,b) F(x) dx will mean Integrate x from a --> b.

We will get:

I = ∫(∞, 0) -e^(-u) du/u^(1/2)

= - ∫(∞, 0) e^(-u) du/u^(1/2)

= ∫(0, ∞) e^(-u) du/u^(1/2)

Note we first pulled the minus sign out of the integral.

Then we changed the limits of intergration from Infinity --> 0 to 0 --> Infinity which eliminates the minus sign.

This is using -( ∫(a, b) f(x) dx ) = ∫(b, a) f(x) dx.

We can now say I = ∫(0, ∞) u^(-1/2) * e^(-u) du

by writing u^(-1/2) in the numerator instead of writing u^(1/2) in the denominator.

This is the same idea as a^(-1) = 1/a

Note using -1/2 = 1/2 -1 we can write

I = ∫(0, ∞) u^(1/2 - 1) * e^(-u) du

Now Γ(z) = ∫(0, ∞) u^(z - 1) * e^(-u) du

Hence I = Γ(1/2)

and we know that Γ(1/2) = Sqrt(π).

Thus the answer is Sqrt(π)

Hope that helps.

Please let me know if you have any questions.