Integral Evaluation Strategy

To evaluate the integral $ \int_0^1 \frac{dt}{\sqrt{(-\log_e t)}} $, we can use a substitution combined with the definition of the Gamma function.

Step 1: Substitution

Let $ u = -\log_e t $. This implies $ t = e^{-u} $. Differentiating with respect to $u$, we get $ dt = -e^{-u} du $.

Now, we need to change the limits of integration:

• When $ t = 1 $, $ u = -\log_e 1 = 0 $.
• As $ t \to 0^+ $, $ \log_e t \to -\infty $, so $ u = -\log_e t \to \infty $.

Substituting these into the integral gives:

Step 2: Adjusting Limits and Simplifying

We can reverse the limits of integration by changing the sign:

Rewrite $ \sqrt{u} $ as $ u^{1/2} $:

Step 3: Applying the Gamma Function

Recall the definition of the Gamma function: $ \Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx $.

Our integral is in the form $ \int_0^\infty u^{(1/2)-1} e^{-u} du $. By comparing this with the Gamma function definition, we can see that $ z = 1/2 $.

Therefore, the value of the integral is $ \Gamma(1/2) $.

Step 4: Final Result

It is a standard result that $ \Gamma(1/2) = \sqrt{\pi} $.

Thus, the value of the integral $ \int_0^1 \frac{dt}{\sqrt{(-\log_e t)}} $ is $ \sqrt{\pi} $.