Finding minimum value of an integral For $0<\lambda<0.5$ we define the integral:
$$ \int_{0}^{\infty}{\frac{1}{1+y}\cdot y^{-2\lambda}}dy $$ We have to find the minimum value of this integral as $\lambda$ varies in the interval $(0,0.5)$.
What I did was first put $y=1/x$ to reduce it to $$\int_{0}^{\infty}\frac{x^{2\lambda-1}}{1+x}dx $$ Now I defined $$I(\alpha)=\int_{0}^{\infty}e^{-\alpha(x+1)}\frac{x^{2\lambda-1}}{1+x}dx $$ and differenciate w.r.t $\alpha$. Since $I(\infty)=0$, I integrated the expression from $0$ to $\infty$. The end result was that $$I(0)=(-2\lambda)!(2\lambda-1)!$$ Now how do I find the minimum value of the above expression. I do not know how to deal with $\Gamma$ or Beta functions... according to the book, the value is $$\pi\over \sin(2\pi \lambda)$$ How are these two equivalent?

A Second Approach Not Using Beta or Gamma Functions
The function
$$
f(\lambda)=\int_0^\infty\frac{y^{-2\lambda}}{1+y}\,\mathrm{d}y\tag1
$$
is convex:
$$
\begin{align}
f''(\lambda)
&=\int_0^\infty\frac{4\log(y)^2y^{-2\lambda}}{1+y}\,\mathrm{d}y\tag{2a}\\[6pt]
&\ge0\tag{2b}
\end{align}
$$
Furthermore, by substituting $y\mapsto1/y$, we see that
$$
\begin{align}
f'(1/4)
&=\int_0^\infty\frac{-2\log(y)\,y^{-1/2}}{1+y}\,\mathrm{d}y\tag{3a}\\
&=\int_0^\infty\frac{2\log(y)\,y^{-1/2}}{1+y}\,\mathrm{d}y\tag{3b}\\[6pt]
&=0\tag{3c}
\end{align}
$$
Therefore, the minimum occurs at $\lambda=\frac14$. To evaluate at $\lambda=\frac14$, we can substitute $y\mapsto y^2$:
$$
\begin{align}
\int_0^\infty\frac{y^{-1/2}}{1+y}\,\mathrm{d}y
&=2\int_0^\infty\frac{\mathrm{d}y}{1+y^2}\tag{4a}\\
&=\left.2\arctan(y)\vphantom{\int}\right]_0^\infty\tag{4b}\\[5pt]
&=\pi\tag{4c}
\end{align}
$$

Original Answer Showing How to Apply the Beta Function
$$
\begin{align}
\int_0^\infty\frac{y^{-2\lambda}}{1+y}\,\mathrm{d}y
&=\Gamma(1-2\lambda)\Gamma(2\lambda)\tag1\\
&=\pi\csc(2\pi\lambda)\tag2
\end{align}
$$
Explanation:
$(1)$: Beta function integral
$(2)$: Euler's reflection formula
What is the maximum of $\sin(2\pi\lambda)$ on $\left[0,\frac12\right]$?

How To Evaluate the Integral
This answer uses contour integration to evaluate a generalization of the integral, which shows that  the integral is $\pi\csc(2\pi\lambda)$. It also derives Euler's Reflection Formula.
