Finding minimum value of an integral

Question

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?

Answer 1

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 $y\mapsto y^2$: $$ \begin{align} \int_0^\infty\frac{y^{-1/2}}{1+y}\,\mathrm{d}y &=2\int_0^\infty\frac{\mathrm{d}y}{1+y^2}\tag{4a}\\ &=\left.2\arctan(y)\vphantom{\int}\right]_0^\infty\tag{4b}\\[5pt] &=\pi\tag{4c} \end{align} $$

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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 $\left[0,\frac12\right]$?

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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.