If $g(n)=\int_0^{n^2+n+1}e^{\frac x2-[\frac x2]}(\frac x2-[\frac x2])d(x-[x])$, then find minima/maxima for $g(n)$

Question

The following question is taken from the practice set of JEE.

Question:

If $g(n)=\int_0^{n^2+n+1}e^{\frac x2-[\frac x2]}(\frac x2-[\frac x2])d(x-[x]),$ where $[x]$ is the greatest integer function and $n\in\mathbb N$, then $g(n)$

(1) has minimum value as $\frac14+\sqrt e$

(2) has maximum value as $3-\sqrt e$

(3) has minimum value as $\frac34-\sqrt{\frac e4}$

(4) none of these

My Attempt:

If $n\in\mathbb N$, then $n^2+n+1$ always increases with minimum value $3$.

Also, $x-[x]=\{x\}$, where $\{x\}$ is fractional part function.

If, in the integral, I had $dx$ then I would have written

$g'(n)=e^{\{n^2+n+1\}}(\{n^2+n+1\})(2n+1)$

But now, we have $d\{x\}$ in integral. I wonder how to use the leibnitz differentiation now.

Also, when $x\in(0,1)$ then in the integral, we can write $d(x)$

Also when $x\in(1,2)$ then in the integral, we can write $d(x-1)$

Edit:

Since $\{x\}$ has period $1$,

So, $g(n)=(n^2+n+1)\int_0^1e^{\frac x2}\frac x2 dx$

Or, do I need to consider the periodicity of $\{\frac x2\}$?

Then $g(n)=2(n^2+n+1)\int_0^1e^{\frac x2}\frac x2 dx$

And $\int_0^1e^{\frac x2}\frac x2 dx=2-\sqrt e$

In either case, the answer comes out to be option (4).

But I just wonder which case to consider.

Edit$2$:

$\int_0^{3π}|\sin x|dx=3\int_0^π\sin xdx$ because period of $|\sin x|$ is π.

I have used this concept above.

For, $g'(n)$, I used Leibnitz integral rule.

Edit$3$:

I am taking a simple case,

$I=\int_0^4\{\frac x2\}d\{x\}$

I wonder if I should consider the periodicity of $\{x\}$ or $\{\frac x2\}$

In one case, $I=4\int_0^1\frac x2dx$

In other case, it's $I=2\int_0^2\frac x2d\{x\}$

I wonder which one to take.

Answer 1

Let $f(x)=x-[x]=x-n$, $\ \ x\in [n,n+1)$

then $f(\frac{x}{2})=\frac{x}{2}-n, \ \ x\in[2n,2n+2)$

$$f’(x)=1(x\not=n)$$

$$\int_0^{n^2+n+1}f(\frac{x}{2})e^{f(\frac{x}{2})}d(x-[x])=\int_0^{n^2+n+1}f(\frac{x}{2})e^{f(\frac{x}{2})}d(x)$$

$$=\sum_{k=0}^{\frac{n^2+n-2}{2}}\int_{2k}^{2k+2}(\frac{x}{2}-k)e^{(\frac{x}{2}-k)}dx+\int_{n^2+n}^{n^2+n+1}(\frac{x}{2}-\frac{n^2+n}{2})e^{(\frac{x}{2}-\frac{n^2+n}{2})}dx$$

$$=2\sum_{k=0}^{\frac{n^2+n-2}{2}}\int_{2k}^{2k+2}(\frac{x}{2}-k)e^{(\frac{x}{2}-k)}d(\frac{x}{2}-k)+2\int_{n^2+n}^{n^2+n+1}(\frac{x}{2}-\frac{n^2+n}{2})e^{(\frac{x}{2}-\frac{n^2+n}{2})}d(\frac{x}{2}-k)$$

$$=2\sum_{k=0}^{\frac{n^2+n-2}{2}}(\frac{x}{2}-k-1)e^{(\frac{x}{2}-k)}|_{2k}^{2k+2}+2(\frac{x}{2}-\frac{n^2+n}{2}-1)e^{(\frac{x}{2}-\frac{n^2+n}{2})}|_{n^2+n}^{n^2+n+1}$$

$$=2\sum_{k=0}^{\frac{n^2+n-2}{2}}1+2(-\frac{\sqrt e}{2}+1)$$

$$=n^2+n+2-\sqrt e$$

Answer 2

The Leibniz integral rule can only be used when the integrand function is continuous as if it isn't, then the anti-derivative will not be differentiable in $\mathbb R$ as it will be a piecewise function.

For example, consider the integral $F(k)=\displaystyle \int_0^k \{x\}\,\mathrm dx, k\in\mathbb R$. This definite integral can only be solved graphically as $\{x\}$ is discontinuous $\forall\, x\in\mathbb Z$. So, we get $$F(k)=\int_0^k\{x\}\,\mathrm dx=[k]\int_0^1x\,\mathrm dx+\int_{[k]}^{k}x-[k]\,\mathrm dx=\frac{[k]+\{k\}^2}2$$ Note that the anti-derivative is non-differentiable for integral values of $x$ as the LHD is $1$ but the RHD is $0$. So, the Leibniz integral rule fails to work when $k$ is an integer. In fact, it only works for $k\in[0,1)$. This can be seen by writing the anti-derivative as an infinite combination of the following functions: $$F(k)=\begin{cases}\cdots\\ \frac{-1+(k+1)^2}2, k\in[-1,0)\\ \frac{k^2}2, k\in[0,1)\\ \frac{1+(k-1)^2}2, k\in[1,2)\\ \cdots\end{cases}$$ So, if we differentiate $F(k)$, the constant term, i.e., $\frac{[k]}2$ vanishes except for $k\in[0,1)$(since it is already zero) which explains why the Leibniz integral rule can't be used for intervals of $k$ other than $[0,1)$: $$F'(k)=\begin{cases}\cdots\\ k+1,k\in[-1,0)\\ k,k\in[0,1)\\ k-1, k\in[1,2)\\ \cdots\end{cases}$$

Hence, the integral in concern should be first simplified by exploiting the integrand's periodicity, then we can proceed by evaluating it as an indefinite integral and plugging in the bounds. The aforementioned property is:

If a periodic function $f(x)$ is continuous in $(a+mT, b+nT)$ with period $T$, then for $a,b\in\mathbb R$ amd $m,n\in\mathbb z$ $$\int_{a+mT}^{b+nT}f(x)\,\mathrm dx=\int_a^bf(x)\,\mathrm dx+(n-m)\int_0^Tf(x)\,\mathrm dx$$ Proof: $$\begin{align}\int_{a+mT}^{b+nT}f(x)\,\mathrm dx&=\int_{a+mT}^{a+nT}f(x)\,\mathrm dx+\int_{a+nT}^{b+nT}f(x)\,\mathrm dx\\&\overset{t=x-mT, u=x-nT}{=}\int_{a}^{a+(n-m)T}f(t)\,\mathrm dt+\int_a^bf(u)\,\mathrm du\\&=\int_{a}^{(n-m)T}f(x)\,\mathrm dx+\underbrace{\int_{(n-m)T}^{a+(n-m)T}f(x)\,\mathrm dx}_{=\displaystyle\int_0^af(x)\,\mathrm dx}+\int_a^bf(x)\,\mathrm dx\\&=\int_a^bf(x)\,\mathrm dx+(n-m)\int_0^Tf(x)\,\mathrm dx\end{align}$$

Since this property is solely dependent on the integrand's periodicity, we will be considering the periodicity of $\{\frac{x}2\}$ and not $\{x\}$ for evaluating $g(n)$.

Similarly, the correct simplication of $\displaystyle\int_0^4\left\{\frac{x}2\right\}\,\mathrm d\{x\}$ is $2\displaystyle\int_0^2\left\{\frac{x}2\right\}\,\mathrm d\{x\}$ and not $4\displaystyle\int_0^1\left\{\frac{x}2\right\}\,\mathrm d\{x\}$.

Now, noting that $\mathrm d\{x\}=\mathrm dx$ for non-integral values of $x$, we need to split $g(n)$ into integrals such that in each integral, the integrand is continuous in the open interval defined by its bounds:
$$\begin{align}g(n)&=\int_0^2e^{\left\{\frac{x}2\right\}}\left\{\frac{x}2\right\}\mathrm dx+\int_{0+1\times2}^{2+1\times2}e^{\left\{\frac{x}2\right\}}\left\{\frac{x}2\right\}\mathrm dx+\int_{0+2\times2}^{2+2\times2}e^{\left\{\frac{x}2\right\}}\left\{\frac{x}2\right\}\mathrm dx+\dots+\int_{0+\frac{n^2+n-2}2\times2}^{2+\frac{n^2+n-2}2\times2}e^{\left\{\frac{x}2\right\}}\left\{\frac{x}2\right\}\mathrm dx+\int_{\frac{n^2+n}2\times2}^{\frac{n^2+n}2\times2+1}e^{\left\{\frac{x}2\right\}}\left\{\frac{x}2\right\}\mathrm dx\\&=\frac{n^2+n}2\int_0^2e^{\left\{\frac{x}2\right\}}\left\{\frac{x}2\right\}\mathrm dx+\int_0^1e^{\left\{\frac{x}2\right\}}\left\{\frac{x}2\right\}\mathrm dx\\&=\frac{n^2+n}2\int_0^2e^{\frac{x}2}\frac{x}2\mathrm dx+\int_0^1e^{\frac{x}2}\frac{x}2\mathrm dx\\&=(n^2+n)\int_0^1te^t\mathrm dt+\int_0^\frac12te^t\mathrm dt\\&=n^2+n+2-\sqrt e\end{align}$$

Hence, the minimum value of $g(n)$ is $4-\sqrt e$ and hence, option $(\text D)$ is correct.

Answer 3

$$g(n)=\sum_{k=0}^{n^2+n}\int_{k}^{k+1} e^{\frac x2-[\frac x2]}(\frac x2-[\frac x2])d(x-[x])$$

Since $[x] = k$ is constant within each interval $(k, k+1)$, then $d[x] = 0$, so:

$$g(n)=\sum_{k=0}^{n^2+n}\int_{k}^{k+1} e^{\frac x2-[\frac x2]}(\frac x2-[\frac x2]) ~dx$$

Let $u = \frac{x}{2} \implies x = 2u \implies dx = 2 ~du$

$$g(n)=2\sum_{k=0}^{n^2+n}\int_{k/2}^{(k+1)/2} e^{u-[u]}(u-[u]) ~du$$ $$=2\left(\sum_{\text{k even}}\int_{k/2}^{(k+1)/2} e^{u-[u]}(u-[u]) ~du+\sum_{\text{k odd}}\int_{k/2}^{(k+1)/2} e^{u-[u]}(u-[u]) ~du\right)$$

Let $v = u - [u]$. If $k$ is even, then $[u] = \frac{k}{2}$, so $v = u - \frac{k}{2}$. If $k$ is odd, then $[u] = \frac{k-1}2$, so $v = u - \frac{k}{2} + \frac{1}{2}$. In both cases, $du = dv$.

$$g(n)=2\sum_{\text{k even}}\int_{0}^{1/2} ve^{v} ~dv+2\sum_{\text{k odd}}\int_{1/2}^{1} ve^{v} ~dv$$ $$=2\sum_{\text{k even}}[(v-1)e^v]_{0}^{1/2}+2\sum_{\text{k odd}}[(v-1)e^v]_{1/2}^{1}$$ $$=2\sum_{\text{k even}}(\frac{-1}{2}e^{1/2} + 1)+2\sum_{\text{k odd}}\frac{1}{2}e^{1/2}$$ $$=\sum_{\text{k even}}(-e^{1/2} + 2)+\sum_{\text{k odd}}e^{1/2}$$

Since each summand is a constant, the sum is just that constant times the number of items being summed. Note that there are $n^2+n+1$ values of $k$, of which $\frac{n^2+n+2}{2}$ are even and $\frac{n^2+n}{2}$ are odd, so:

$$g(n)=\frac{n^2+n+2}{2}(-\sqrt{e} + 2)+\frac{n^2+n}{2}\sqrt{e}$$ $$\boxed{g(n)= n^2+n+2 - \sqrt{e}}$$

If $n$ were an arbitrary real number, then this would be the equation of a parabola with a vertex at $(-\frac{1}{2}, \frac{11}{4} - \sqrt{3})$. But since we are given that $n \in \mathbb{N}$, then the minimum occurs at eitehr $(0, 2-\sqrt{e})$ or $(1, 4-\sqrt{e})$, depending on whether $\mathbb{N}$ is defined to include or exclude 0. In either case, the minimum value isn't one of the answer choices, so “none of these” is the correct answer.