A peculiar Riemann Sum

Question

Im trying to express the following limit as a Riemann Sum

$$ \lim_{ n \to \infty} \sum_{i=1}^{n} \dfrac{ n^3 (2i-3) }{ (n^2- \sqrt{2n^4 + i^2n^2} + in )(2n^2-i^2)^{3/2} } $$

over the interval $[2,3]$

Clearly, we want $\dfrac{1}{n}$ out of the sum as to take $\Delta x$ in account. We can divide both sides by $\dfrac{1}{n^5}$ and we obtain the

$$ \lim_{ n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \dfrac{ \dfrac{(2i-3)}{n} }{ (n^2- \sqrt{2n^4 + i^2n^2} + in )(2n^2-i^2)^{3/2} } \cdot \frac{1}{\dfrac{1}{n^2 \cdot n^3}}$$

Which reduces to

$$ \lim_{ n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \dfrac{ \dfrac{(2i-3)}{n} }{ \left( 1+ \dfrac{i}{n} - \sqrt{ 2 + \dfrac{i^2}{n^2} } \right)\left(2- \left( \dfrac{i}{n} \right)^2 \right) } $$

Now, this is almost done as we want our height x-value to be $x = 2 + \dfrac{i}{n}$. However, here is where I get stuck and I ask for some help. The term $\dfrac{2i-3}{n}$ in the numerator is the only term that is problematic here. How would we handle this?

Answer 1

According to Riemann sum

$$\int_{a} ^{b} f(x) \, dx=\lim_{n\to\infty} \frac{b-a} {n} \sum_{i=1}^{n}f\left(a+i\cdot\frac {b-a} {n} \right)$$

we need to express the sum as a function of $\left(2+\frac{i}{n}\right)$ that is

$$\lim_{ n \to \infty} \frac{1}{n}\sum_{i=1}^{n} \frac{ 2\left(2+\frac{i}{n}\right)-4+\frac 3 n }{ \left( \left(2+ \frac{i}{n}\right)-1 - \sqrt{ \left(2 + \frac{i}{n}\right)^2-4\left(2+\frac in\right)+6 } \right)\left(-2+4\left(2+\frac i n\right)- \left(2+ \frac{i}{n} \right)^2 \right) }=$$

$$=\int_2^3\frac{2x-4}{(x-1-\sqrt{x^2-4x+6})(-2+4x-x^2)}dx$$

which is an improper integral since $x-1-\sqrt{x^2-4x+6}=0$ for $x=\frac 5 2$.

Refer also to

• Perfect understanding of Riemann Sums

Answer 2

As @conditionalMethod pointed out, for $n$ even, one of the denominators is zero. If we exclude $2i-n=0$ in the summation, the limit exits and equals to some proper Riemann integral. In other words, usual Riemann integral is enough to deal with the problem, and improper integral is not necessarily introduced. Also, from @conditionalMethod's comments on @user's answer, we may need to deal with the term $\frac{3}{n}$ carefully.

Let us prove that \begin{align} &\lim_{n\to \infty} \sum_{i=1, \ 2i-n\ne 0}^n \frac{n^3(2i-3)}{(n^2-\sqrt{2n^4+i^2n^2} + in)(2n^2-i^2)^{3/2}}\\ =\ & \int_0^1 \frac{1+x+\sqrt{2+x^2}}{(2-x^2)^{3/2}} \mathrm{d}x \\ & + \int_0^1 \frac{1}{2(1-2x)} \Big(\frac{1 + (1-x) + \sqrt{2 + (1-x)^2}}{(2-(1-x)^2)^{3/2}} - \frac{1 + x + \sqrt{2 + x^2}}{(2-x^2)^{3/2}} \Big)\mathrm{d} x. \end{align} Remark: The second integral is a proper Riemann integral rather than an improper integral, since the integrand is continuous on $[0, 1]$ (it is obvious if it is rewritten to reduce $1-2x$, although the expression hence becomes complicated).

Proof: First, we have \begin{align} &\sum_{i=1, \ 2i-n\ne 0}^n \frac{n^3(2i-3)}{(n^2-\sqrt{2n^4+i^2n^2} + in)(2n^2-i^2)^{3/2}}\\ =\ & \sum_{i=1, \ 2i-n\ne 0}^n \frac{n^3(2i-n)}{(n^2-\sqrt{2n^4+i^2n^2} + in)(2n^2-i^2)^{3/2}}\\ &\qquad\qquad + \sum_{i=1, \ 2i-n\ne 0}^n \frac{n^3(n-3)}{(n^2-\sqrt{2n^4+i^2n^2} + in)(2n^2-i^2)^{3/2}}\\ =\ & \sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n+i +\sqrt{2n^2+i^2})}{(2n^2-i^2)^{3/2}} + \sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)(n+i+\sqrt{2n^2+i^2})}{(2i-n)(2n^2-i^2)^{3/2}}. \end{align}

Second, we have \begin{align} &\lim_{n\to \infty} \sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n+i +\sqrt{2n^2+i^2})}{(2n^2-i^2)^{3/2}} \\ = \ & \lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n \frac{1 + \frac{i}{n} + \sqrt{2 + (\frac{i}{n})^2}}{(2 - (\frac{i}{n})^2)^{3/2}}\\ = \ & \int_0^1 \frac{1+x+\sqrt{2+x^2}}{(2-x^2)^{3/2}} \mathrm{d}x. \end{align}

Third, since (with substitution $i \rightarrow n-i$) \begin{align} &\sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)(n+i+\sqrt{2n^2+i^2})}{(2i-n)(2n^2-i^2)^{3/2}}\\ =\ & \sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)(n+(n-i)+\sqrt{2n^2+(n-i)^2})}{(2(n-i)-n)(2n^2-(n-i)^2)^{3/2}}\\ &\quad + \frac{(n-3)(1+\sqrt{2})}{2n^2 \sqrt{2}} + \frac{(n-3)(2+\sqrt{3})}{n^2}, \end{align} we have \begin{align} &\sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)(n+i+\sqrt{2n^2+i^2})}{(2i-n)(2n^2-i^2)^{3/2}}\\ =\ & \frac{1}{2}\sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)(n+i+\sqrt{2n^2+i^2})}{(2i-n)(2n^2-i^2)^{3/2}}\\ &\quad + \frac{1}{2} \sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)(n+(n-i)+\sqrt{2n^2+(n-i)^2})}{(n-2i)(2n^2-(n-i)^2)^{3/2}}\\ &\quad + \frac{1}{2} \frac{(n-3)(1+\sqrt{2})}{2n^2 \sqrt{2}} + \frac{1}{2}\frac{(n-3)(2+\sqrt{3})}{n^2}\\ =\ & \sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)}{2(n-2i)}\Big(\frac{n+(n-i)+\sqrt{2n^2+(n-i)^2}}{(2n^2-(n-i)^2)^{3/2}} - \frac{n+i+\sqrt{2n^2+i^2}}{(2n^2-i^2)^{3/2}}\Big)\\ &\quad + \frac{1}{2} \frac{(n-3)(1+\sqrt{2})}{2n^2 \sqrt{2}} + \frac{1}{2}\frac{(n-3)(2+\sqrt{3})}{n^2}\\ =\ & \frac{n-3}{n} \cdot \frac{1}{n} \sum_{i=1, \ 2i-n\ne 0}^n f\big(\frac{i}{n}\big) + \frac{1}{2} \frac{(n-3)(1+\sqrt{2})}{2n^2 \sqrt{2}} + \frac{1}{2}\frac{(n-3)(2+\sqrt{3})}{n^2}\\ =\ & \frac{n-3}{n} \cdot \frac{1}{n} \sum_{i=1}^n f\big(\frac{i}{n}\big) - \frac{n-3}{n^2}f\big(\frac{1}{2}\big)\cdot \mathrm{mod}(n-1,2)\\ &\quad + \frac{1}{2} \frac{(n-3)(1+\sqrt{2})}{2n^2 \sqrt{2}} + \frac{1}{2}\frac{(n-3)(2+\sqrt{3})}{n^2} \end{align} where $\mathrm{mod}(n-1,2) = 1$ is $n$ is even and $\mathrm{mod}(n-1,2) = 0$ is $n$ is odd, and $$f(x) = \frac{1}{2(1-2x)} \Big(\frac{1 + (1-x) + \sqrt{2 + (1-x)^2}}{(2-(1-x)^2)^{3/2}} - \frac{1 + x + \sqrt{2 + x^2}}{(2-x^2)^{3/2}} \Big).$$ Here we have used the fact that $f(x)$ is continuous on $[0, 1]$ with $f(\frac{1}{2})=\frac{328}{1029}\sqrt{7}$. Thus, we have $$\lim_{n\to \infty} \sum_{i=1, \ 2i-n\ne 0}^n \frac{n(n-3)(n+i+\sqrt{2n^2+i^2})}{(2i-n)(2n^2-i^2)^{3/2}} = \int_0^1 f(x) dx.$$ The desired result follows. This completes the proof.