Secret Spice

Alternate Solution:

Using Maclaurin series, $\frac { 1 }{ 1+x } =1-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }+...$

On converting it into a beautiful summation, $\frac { 1 }{ 1+x } =\sum _{ k=0 }^{ \infty }{ \frac { \Gamma \left( 1+k \right) }{ k! } { \left( -x \right) }^{ k } }$

(You can prove it by using properties of Gamma function)

Now, the integral is: $M\left\{ \frac { 1 }{ 1+x } \right\} \left( a \right)$

Here M{f(x)}(s) is Mellin transform of f(x) over s.

Now using Ramanujan's Master Theorem , we get $\int _{ 0 }^{ \infty }{ \frac { { \left( x \right) }^{ a-1 } }{ 1+x } } dx=\Gamma \left( a \right) \Gamma \left( 1-a \right)$

Since $0<a<1$ , we can use Euler's reflection formula.

Therefore, the above integral comes out to be: $\frac { \pi }{ sin\left( a\pi \right) }$

$I = \displaystyle \int_{0}^{\infty} \dfrac{x^{a-1}}{1+x}dx$

$$ Let $x = \tan^2 \theta$

$$ $dx = 2 \tan \theta \sec^2 \theta d\theta$

$$ $I = \displaystyle2 \times \int_{0}^{\dfrac{\pi}{2}} \tan^{2a-2}\theta \times \tan\theta \sec^2 \theta \dfrac{d\theta}{sec^2 \theta}$

$\therefore I = \displaystyle 2 \times \int_{0}^{\dfrac{\pi}{2}} \sin^{2a-1}(\theta)\cos^{1-2a}(\theta)d\theta$

$$ $\therefore I = \displaystyle 2 \times \dfrac{\beta(a,1-a)}{2}$

$$ $\therefore I = \displaystyle \Gamma(a) \Gamma(1-a)$

$$ Using Euler's reflection formula,

$\Gamma(x) \Gamma(1-x) = \dfrac{\pi}{\sin\pi x}$

$$ $\therefore I = \displaystyle \dfrac{\pi}{\sin\pi a}$

A logical solution to this problem Put a = 1/2 Since the function is positive in given limits and the integral has a closed form in 0 < a < 1 Hence the function cannot have infinite or zero area under the curve in X-Y plane , this leaves us out with only one option possible i.e pi/sin(a*pi).