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:
$ \int_0^1 \frac{dt}{\sqrt{(-\log_e t)}} = \int_{\infty}^0 \frac{-e^{-u} du}{\sqrt{u}} $Step 2: Adjusting Limits and Simplifying
We can reverse the limits of integration by changing the sign:
$ \int_{\infty}^0 \frac{-e^{-u} du}{\sqrt{u}} = \int_0^{\infty} \frac{e^{-u}}{\sqrt{u}} du $Rewrite $ \sqrt{u} $ as $ u^{1/2} $:
$ \int_0^{\infty} \frac{e^{-u}}{u^{1/2}} du = \int_0^{\infty} u^{-1/2} e^{-u} du $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} $.
Important Questions from Definite Integrals
The Legendre polynomials P n(x), n = 0, 1, 2, ..., satisfying the orthogonailty condition \(\int_{{\rm{ - 1}}}^{\rm{1}} {{{\rm{P}}_{\rm{n}}}\left( {\rm{x}} \right){{\rm{P}}_{\rm{m}}}} \left( {\rm{x}} \right){\rm{dx}}\,{\rm{ = }}\,\frac{{\rm{2}}}{{{\rm{2n + 1}}}}{{\rm{\delta }}_{{\rm{nm}}}}\) on the interval [-1, +1], may be defined by the Rodrigues formula P n(x) = \(\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}{\rm{n!}}}}\frac{{{{\rm{d}}^{\rm{n}}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} \right)^{\rm{n}}}\) . The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is
A parametric curve is defined \(x = cos\left(\frac{\Pi t}{2}\right) , Y= sin\left(\frac{\Pi t}{2}\right)\) in the range of \(0\leq t\leq 1\) . It is rotated about X-axis by 360°.
‘What is the area of the surface generated?
if \(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx=Ax+B\log |sin(x-a)|+ C\) where A, B and c are real constants then:
Which of the following is NOT a property of definite integral?
The value of integral \(x\over cos^2 x\) is equal to
