A cot integral limit

Evaluate the limit:

    \[\ell = \lim_{x \rightarrow +\infty} \int_{\frac{1}{x+1}}^{\frac{1}{x}} \cot t^2 \, \mathrm{d}t\]


Recalling the Taylor expansion of \tan around x=0 we have that

    \[\cot x^2 = \frac{1}{x^2} + \mathcal{O}\left ( x^2 \right )\]


    \begin{align*} \int_{\frac{1}{x+1}}^{\frac{1}{x}} \cot t^2 \, \mathrm{d}t &= \int_{\frac{1}{x+1}}^{\frac{1}{x}} \left ( \cot t^2 - \frac{1}{t^2} + \frac{1}{t^2} \right ) \, \mathrm{d}t \\ &=\int_{\frac{1}{x+1}}^{\frac{1}{x}} \left ( \cot t^2 - \frac{1}{t^2} \right ) \, \mathrm{d}t + \int_{\frac{1}{x+1}}^{\frac{1}{x}} \frac{\mathrm{d}t}{t^2} \\ &=\mathcal{O} \left ( \int_{\frac{1}{x+1}}^{\frac{1}{x}} t^2 \right ) + 1 \\ &= \mathcal{O} \left ( \frac{1}{x^3} - \frac{1}{(x+1)^3} \right ) +1 \\ &= 1 +\mathcal{O} \left ( \frac{1}{x^4} \right ) \end{align*}

The limit follows to be 1.

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Definite integral

Let f:[0, 1]\rightarrow \mathbb{R} be a function such that

    \[f(x^2)+\sqrt {x} f \left( x^2\sqrt {x} \right) = e^x \quad \text{forall} \;\; x \in [0, 1]\]

Evaluate the integral \int_0^1 f(x)\, \mathrm{d}x.


We multiply the given equation by x. Thus,

\begin{aligned} f \left(x^2 \right)+\sqrt {x} f \left(x^2\sqrt {x} \right) &= e^x \Leftrightarrow x f \left(x^2 \right)+x\sqrt{x} f\left(x^2\sqrt {x}\right) =x e^x \\ &\Leftrightarrow \int_{0}^{1} x f \left ( x^2 \right )\, \mathrm{d}x + \int_{0}^{1} x \sqrt{x} f \left ( x^2 \sqrt{x} \right )\, \mathrm{d}x = \int_{0}^{1} xe^x \, \mathrm{d}x \end{aligned}

Let \mathcal{J}_1=\int_{0}^{1} x f \left ( x^2 \right )\, \mathrm{d}x and \mathcal{J}_2= \int_{0}^{1} x \sqrt{x} f \left ( x^2 \sqrt{x} \right )\, \mathrm{d}x.

We now deal with the first integral:

    \begin{align*} \int_{0}^{1}x f\left ( x^2 \right )\, \mathrm{d}x &\overset{u=x^2}{=\! =\!=\!} \frac{1}{2}\int_{0}^{1} f(u)\, \mathrm{d}u \end{align*}

As for the second integral we have:

    \begin{align*}\int_{0}^{1} x\sqrt{x} f\left ( x^2 \sqrt{x} \right )\, \mathrm{d}x &\overset{u=x^2 \sqrt{x}}{=\! =\! =\! =\! =\!} \frac{2}{5} \int_{0}^{1} f(u)\, \mathrm{d}u \end{align*}


    \[\int_{0}^{1} f(x) \, \mathrm{d}x = \frac{10}{9}\]

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Offset logarithmic integral inequality

Prove that

    \[\int_2^{e+1} \frac{\mathrm{d}t}{\ln t} < e\]


We have successively:

    \begin{align*} \hspace{-1em}\ln x \leq x-1 &\Rightarrow \ln \frac{1}{x} \leq \frac{1}{x}-1 \\ &\Rightarrow -\ln x \leq \frac{1}{x}-1 \\ &\Rightarrow \ln x \geq 1-\frac{1}{x} \\ &\Rightarrow \frac{1}{\ln x} \leq \frac{1}{1-\frac{1}{x}} \\ &\Rightarrow \int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < \int_2^{e+1} \frac{\mathrm{d}t}{1-\frac{1}{t}} \\ &\Rightarrow \mathbf{\int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < e} \end{align*}

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Limit of geometric mean of binomial coefficients

Let G_n denote the geometric mean of the binomial coefficients

    \[\binom{n}{0}, \; \binom{n}{1}, \; \binom{n}{2}, \; \cdots, \; \binom{n}{n}\]

Prove that \lim \limits_{n \rightarrow +\infty} \sqrt[n]{G_n}=\sqrt{e}.


We note that

    \begin{align*} \sqrt[n]{G_n}&=\exp\left(\frac{1}{n(n+1)}\sum_{k=0}^n\left(\ln n!-\ln k!-\ln((n-k)!)\right)\right)\\ &=\exp\left(\frac{1}{n}\sum_{k=1}^n\ln k-\frac{2}{n(n+1)}\sum_{k=1}^n\sum_{m=1}^k\ln m\right) \end{align*}

On the other hand the following lemma holds:

Lemma: Let f:[M, N] \rightarrow \mathbb{R} be a monotonic function. It holds that


Proof: Due to monotony it holds that f(k+1)\leq \int_k^{k+1}f(x)\,\mathrm{d}x \leq f(k)for k=M,\ldots,N-1. Hence summing over all these values of k we get that

    \[-f(M)\leq\int_M^Nf(x)\,\mathrm{d}x-\sum_{k=M}^Nf(k)\leq -f(N)\]

The result follows.

Applying the above to f(x)=\ln x on [1, k] we get that:

    \[\left|\int_{1}^{k}\ln(x)\,dx-\sum_{m=1}^k\ln m\right|\leq \ln k\]


(1)   \begin{equation*} \sum_{m=1}^k\ln m=\int_{1}^{k}\ln x\,\mathrm{d}x+\mathcal O(\ln k)=k\ln k-k+\mathcal{O}\left(\max\{1,\ln k\}\right) \end{equation*}

Similarly, applying the above to f(x)=x\ln x on [1, n] we get that:

(2)   \begin{equation*} \sum_{k=1}^nk\ln k=\int_{1}^{n}x\ln x\,\mathrm{d}x+\mathcal O(n\ln n)=\frac{n^2\ln n}{2}-\frac{n^2}{4}+\mathcal O(n\ln n) \end{equation*}

The result follows.

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Digamma and Trigamma functions

Let \psi^{(0)} and \psi^{(1)} denote the digamma and trigamma functions respectively. Prove that:

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n + \frac{1}{2} \psi^{(1)}(n) \right ) = 1+ \frac{\gamma}{2} - \frac{\ln 2\pi}{2}\]

where \gamma denotes the Euler – Mascheroni constant.


We begin with the recently discovered identity:

    \begin{align*} \log \Gamma(1+x) &= \frac{\ln 2 \pi -1}{2} -\gamma \left ( x+\frac{1}{2} \right ) \\ &\quad + \frac{2x-1}{2} + \sum_{n=1}^{\infty} \left ( \psi^{(0)}(n+1) -\ln(x+n) +\frac{2x-1}{2(1+n)} \right ) \end{align*}

Letting x=1 we get that

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n +\frac{1}{2n} \right ) = \frac{1+\gamma-\ln 2\pi}{2}\]

Now combining this result here we conclude the exercise.

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