Example 2.3.5; Functional Analysis book by S. Kesavan The author in this example is trying to show that the norm of Hilbert matrix is less than or equal to $\pi$. Hilbert matrix has entries
$$a_{ij}=\frac{1}{i+j+1};1\leq i,j \leq \infty$$ He used the fact that if $\exists$ a sequence of positive real numbers $({p_{i}})$ with $\beta >0$ and $\gamma >0$  satisfying $$\sum_{i=1}^{\infty}a_{ij}p_{i}\leq\beta p_{j};\:\forall j\in\mathbb{N}$$
and
$$\sum_{j=1}^{\infty}a_{ij}p_{j}\leq\gamma p_{i};\:\forall i\in\mathbb{N}$$
Then $A\in L(l_{2})$ and $||A||^{2}\leq \beta\gamma$.
For Hilbert matrix he considered,
$$\sum_{i=0}^{\infty}a_{ij}p_{i} = \sum_{i=0}^{\infty}\frac{1}{(i+1/2+j+1/2)\sqrt{i+1/2}}$$
$$<\int_{0}^{\infty}\frac{dx}{(x+j+1/2)\sqrt{x}}$$
$$=2\int_{0}^{\infty}\frac{dt}{t^{2}+j+1/2}$$
$$=\frac{\pi}{\sqrt{j+1/2}}$$
I am not able to understand the second step of the proof that how he is able to relate summation with integration.
 A: I'd like to make an extended comment on the remark in the OP "I am not able to understand the second step of the proof that how he is able to relate summation with integration."
I had the same difficulty, and resolved it by carefully looking at the replacement of the sum by an integral.
The sum in question is
$$s(j) = \sum_{i=0}^{\infty} a(i,j)\tag{1}$$
where the summand is
$$a(i,j)=\frac{1}{\sqrt{i+\frac{1}{2}} (i+j+1)}\tag{2}$$
A  modified summand which will become an integrand subsequently is
$$b(x,j) = a(i\to x-\frac{1}{2},j)=\frac{1}{\sqrt{x}(x+j+\frac{1}{2})}\tag{3}$$
Now consider the graph

The trick is to look at the half integer points. Because the second derivative $\frac{\partial^2 b(x,j)}{\partial x^2} >0$ the integral between consecutive integer points is greater than the area of the corresponding rectangular strip. To make this evident, compare the areas under the blue curve and those under the yellow curve in the intervals from $i=1$ to $i=\frac{3}{2}$ with that in the interval from $i=\frac{3}{2}$ to $i=2$. We see that the area of "triangle" in the first interval is slightly gerater than that of the second "triangle":
Hence follows the inequality.
The integral is then calculated from the formula
$$\int_0^\infty \frac{1}{\sqrt{x}(x+j+a)} = \frac{\pi}{\sqrt{a+j}}\tag{4}$$
Discussion
Because in our case we have $a=\frac{1}{2}$ we find
$$s(j) < \frac{\pi}{\sqrt{\frac{1}{2}+j}}\tag{5}$$
but numerical evidence shows that even
$$s(j) < \frac{\pi}{\sqrt{1+j}}\tag{6}$$
I was not able to find a proof of the stronger inequality $(6)$. So it remains as an interesting open problem.
We can also easily find a lower bound of $s(j)$ by just integrating
$$s(j) > \int_0^{\infty } a(x,j) \, dx = \frac{2 \cos ^{-1}\left(\frac{1}{\sqrt{2} \sqrt{j+1}}\right)}{\sqrt{j+\frac{1}{2}}}\tag{7}$$
A: I think the sums starting from $i=0$ is a mistake. They should start from $i=1$. Now $\int_{i-1+\frac 1 2}^{i+\frac 1 2} \frac 1 {(x+j+\frac 1 2)\sqrt x } dx > \frac 1 {(i+\frac 1 2 +j+\frac 1 2) \sqrt {i+\frac  1 2}}$. Summing over $i$ from $1$ to $\infty$ gives the inequality. 
A: The summation considers only positive integer values for $i$, while the integral considers all positive real values of $i$.  Since all the other terms in the summation are positive, it follows that the summation can be no bigger than the integral.
Consider, just to make it easier to see, the first three terms:
$$\sum_{i=0}^2 \frac{1}{(i+1/2+j+1/2)\sqrt{i+1/2}} = \frac{1}{(1+j)\sqrt{3/2}} + \frac{1}{(2+j)\sqrt{5/2}} + \frac{1}{(3+j)\sqrt{7/2}} $$
Then an upper-bounding integral for this is:
$$\int_0^3 \frac{dx}{(x+j+1/2)\sqrt{x}} \geq \frac{1}{(j+1)\sqrt{1/2}} + \frac{1}{(j+2)\sqrt{3/2}} + \frac{1}{(j+3)\sqrt{5/2} } $$
where we've established the inequality by taking just three points from the domain of integration ($x=1/2,\ 3/2$ and $5/2$).  Since each term of this can be directly compared with the sum, and each term from the integral is larger, the inequality follows, as adding the in the rest of the domain of integration can only increase the value.
