How to evaluate $\int_{0}^{10} x^{2}d\{x+\frac{1}{2}\}$?

Question

How to evaluate the integral $$\int_{x=0}^{x=10}x^{2}d\left\{x+\frac{1}{2}\right\}$$

This is one another integral which I am having doubt. This integral is also from MIT Quarterfinals $2025$.

My Thoughts :

First of all it took me $2$ days to think for this integral.

But still I thought of only $4$ steps.

Let me explain my thoughts.

I assumed the given integral as $I$. That is I got

$$I=\int_{x=0}^{x=10}x^{2}d\left\{x+\frac{1}{2}\right\}$$

Now I wrote this given integral in the form of

$$I=\int_{x=0}^{x=10}x^{2}dx×\frac{d}{dx}\left(\left\{x+\frac{1}{2}\right\}\right)$$

I really don't know whether this step is correct or wrong.

Now the next step is that I wrote $$\left\{x+\frac{1}{2}\right\}$$ as $$\left(x+\frac{1}{2}\right)-\lfloor{x+\frac{1}{2}\rfloor}$$

Now we can easily say that

$\frac{d}{dx}\left(\left\{x+\frac{1}{2}\right\}\right)=\frac{d}{dx}\left((x+\frac{1}{2})-\lfloor{x+\frac{1}{2}\rfloor}\right)$

Now it is very clear to me that $$\frac{d}{dx}\left(\left\{x+\frac{1}{2}\right\}\right)=1$$ because $\lfloor{x+\frac{1}{2}\rfloor}$ is a constant and its derivative with respect to $x$ will result in $0$.

Therefore finally the integral is turning out as $$\int_{x=0}^{x=10}x^{2}dx$$ which yields the answer to be $\frac{1000}{3}$

But the correct answer mentioned is $\frac{5}{6}$.

Please rectify my mistake.

Answer 1

This integral must be interpreted as a Riemann-Stieltjes integral. Toward that end, note that $\{x+\frac12\}$ is a function which is differentiable with derivative $1$ whenever $x +\frac12 \not\in \mathbb{Z}$, and at the points where $x+\frac12\in\mathbb{Z}$, it has a jump of $-1$.

Together, this means $$\begin{align*} \int_0^{10} x^2\,\mathrm{d}\{x+\frac12\} &= \int_0^{10} x^2 \,\mathrm{d}x - \sum_{0 < x < 10 \\ x+\frac12 \in \mathbb{Z}}x^2 \\ &= \frac{1000}{3} - \sum_{n=0}^9(n+\frac12)^2 \\ &= \frac56 \end{align*}$$

Answer 2

Probably a little advance and even too slow for a Integration bee constest, but here is another solution based on integration by parts.

The functions $G(t)=\{t+\frac12\}$ and $F(t)=t^2$ are right-continuous and of local bounded variation. Integration by parts gives \begin{align} \int_{(a,b]}F(t)\,dG(t)&=F(b)G(b)- F(a)G(a)-\int_{(a,b]}G(t-)\,dF(t)\\ &=F(b)G(b)- F(a)G(a)-\int_{(a,b]}G(t)\,dF(t)\\ &=b^2\{b+\tfrac12\}-a^2\{a+\tfrac12\}-\int^b_a \{t+\tfrac12\} 2t\,dt \end{align} For $a=0$ and $b=10$ this is \begin{align} \int_{(0,10]}t^2\,d\{t+\tfrac12\}=50-2\int_{(0,10]}\{t+\tfrac12\}t\,dt \end{align}

The integral on the right-hand-side is \begin{align} \int_{(0,10]}\{t+\tfrac12\}t\,dt&=\int_{(\frac12,\frac{21}{2}]}\{t\}(t-\tfrac12)\,dt\\ &=\int^1_{1/2}t(t-\tfrac12)\,dt+\int^{\frac{21}{2}}_{10} (t-10)(t-\tfrac12)\,dt +\sum^9_{n=1}\int^{n+1}_n(t-n)(t-\tfrac12)\,dt\\ &=\frac43+\int^{10}_1\Big(t^2-\tfrac{t}{2}\Big)\,dt-\sum^9_{n=1}n^2=\frac{295}{12} \end{align}

Thus,

\begin{align} \int_{(0,10]}\{t+\tfrac12\}t\,dt=50-2\frac{295}{12}=\frac56 \end{align}

Answer 3

$$\int_0^{10}x^2d\{x+\frac12\}=\int_0^{10}x^2\frac{d\{x+\frac12\}}{dx}dx$$Where the derivative,

$$\frac{d\{x+\frac12\}}{dx}=\begin{cases}1, \text{if } x\neq n-\frac12\\-\infty , \text{if } x=n-\frac12\end{cases} $$Thus , $$I=\int_0^{10}x^2dx+\sum_{n=1}^{10}\int_0^{10}x^2\left(-\delta\left(n-\frac12\right)\right)dx=\int_0^{10}x^2dx-\sum_{n=1}^{10}\left(n-\frac12\right)^2=\frac56.$$Where , $\delta(\cdot)$ is the delta Dirac function.