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Definite logarithmic integral

Let 0<\alpha<\beta. Evaluate the integral:

    \[\mathcal{J}=\int_\alpha^\beta \frac{\ln x}{(x+\alpha)(x+\beta)}\, \mathrm{d}x\]

Solution

We have successively:

    \begin{align*} \mathcal{J} &= \int_{\alpha}^{\beta} \frac{\ln x}{\left ( x+\alpha \right )\left ( x+\beta \right )}\, \mathrm{d}x \\ &\!\!\!\!\!\!\!\overset{x \mapsto \alpha \beta/x}{=\! =\! =\! =\! =\!=\!} \int_{\alpha}^{\beta} \frac{\ln \alpha \beta- \ln x}{\left ( x + \alpha \right ) \left ( x + \beta \right )} \, \mathrm{d}x\\ &=\frac{1}{2} \int_{\alpha}^{\beta} \frac{\ln \alpha \beta}{\left ( x+\alpha \right )\left ( x + \beta \right )} \, \mathrm{d}x \\ &=\frac{\ln \alpha \beta}{\beta-\alpha} \ln \left ( \frac{(\alpha+\beta)^2}{4\alpha \beta} \right ) \end{align*}

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Definite parametric integral

Let 0<a<b. Evaluate the integral

    \[\mathcal{J} = \int_{a}^{b} \frac{e^{x/a}-e^{b/x}}{\sqrt{abx+ x^3}} \, \mathrm{d}x\]

Solution

The key substitution is x \mapsto \frac{ab}{u}. Applying it we see that

    \begin{align*} \sqrt{abx+ x^3} &\overset{x \mapsto ab/u}{=\! =\! =\! =\! =\!} \sqrt{\frac{a^2b^2}{u} +\frac{a^3b^3}{u^3} } \\ &=\sqrt{\frac{a^2b^2u^2}{u^3} + \frac{a^3b^3}{u^3}} \\ &=\sqrt{\frac{a^2b^2 \left ( u^2+ab \right )}{u^2 \cdot u}} \\ &=\frac{ab}{u} \sqrt{\frac{u^2+ab}{u}} \end{align*}

Thus,

    \begin{align*} \mathcal{J} &= \int_{a}^{b} \frac{e^{x/a}-e^{b/x}}{\sqrt{abx+ x^3}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{x=ab/u}{=\! =\! =\! =\! =\! =\!} ab\int_{a}^{b} \frac{1}{u^2} \cdot \left ( e^{b/u} - e^{u/a} \right ) \cdot \frac{u}{ab} \cdot \frac{\sqrt{u}}{\sqrt{u^2+ab}} \, \mathrm{d}u \\ &=\int_{a}^{b} \frac{e^{b/u}-e^{u/a}}{\sqrt{abu + u^3}} \, \mathrm{d}u \\ &= - \int_{a}^{b} \frac{e^{u/a}-e^{b/u}}{\sqrt{abu+u^3}} \, \mathrm{d}u \\ &= -\mathcal{J} \end{align*}

Thus , \mathcal{J}=0.

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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.

Solution

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*}

Hence,

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

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Offset logarithmic integral inequality

Prove that

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

Solution

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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A Riemann sum II

Evaluate the limit:

    \[\ell =\lim_{n \rightarrow + \infty}\left(\ln\sqrt[n+1]{\frac{n+1}{n}}+\ln\sqrt[n+2]{\frac{n+2}{n}}+\cdots+\ln\sqrt[3n]{\frac{3n}{n}}\right)\]

Solution

We have successively:

    \begin{align*} \ln \prod_{k=1}^{2n} \left(\frac{n+k}{n}\right)^{\frac{1}{n+k}} &= \sum_{k=1}^{2n} \ln \left(\frac{n+k}{n}\right)^{\frac{1}{n+k}} \\ &= \sum_{k=1}^{2n}\frac{1}{n+k} \ln \left(1+\frac{k}{n}\right) \\ &= \frac{1}{n}\sum_{k=1}^{2n}\frac{1}{1+\frac{k}{n}} \ln \left(1+\frac{k}{n}\right) \\ &\rightarrow \int_{0}^{2} \frac{\ln \left ( 1+x \right )}{1+x} \, \mathrm{d}x \\ &= \frac{\ln^2 3}{3} \end{align*}

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