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Δημοσιεύτηκε: **Κυρ Νοέμ 03, 2019 10:54 am**

Έστω $\alpha, \beta>0$ με $\alpha \neq \beta$ . Να δειχθεί ότι:

$\displaystyle{\int_{0}^{\infty} \frac{1-x^2}{\left ( \alpha x +\beta \right )^2} \frac{\ln \left ( 1+x \right )}{\left ( \beta x + \alpha \right )^2} \, \mathrm{d}x = \frac{1}{\alpha \beta \left ( \alpha^2 - \beta^2 \right )} \ln \frac{\beta}{\alpha}}$

Δημοσιεύτηκε: **Κυρ Μάιος 08, 2022 5:30 pm**

Έστω $\mathcal{J}$ το δοθέν ολοκλήρωμα. Τότε,

$\displaystyle{\begin{aligned} \mathcal{J}&=\int_0^\infty \frac{(1-x^2)\ln(1+x)}{(\alpha x+\beta)^2(\beta x + \alpha)^2} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{x\to\frac{1}{x}}{=\! =\! =\! =\! =\!} -\int_0^\infty \frac{(1-x^2)(\ln(1+x)-\ln x)}{(\alpha x+\beta)^2(\alpha+\beta x)^2} \, \mathrm{d}x \\ &=\frac{1}{2} \int_0^\infty \frac{(1-x^2)\ln x}{(\alpha x+\beta)^2(\alpha+\beta x)^2} \, \mathrm{d} x\\ &=\frac{1}{2\left (\alpha^2-\beta^2 \right )}\int_0^\infty \frac{\ln x}{(\alpha x+\beta)^2} \, \mathrm{d} x- \\ &\quad \quad - \frac{1}{2\left (\alpha^2-\beta^2 \right )}\int_0^\infty \frac{\ln x}{(\alpha+\beta x)^2} \, \mathrm{d}x \end{aligned}}$
Παρατηρούμε ότι

$\displaystyle{\int_{0}^{\infty} \frac{\ln x}{\left ( \alpha + \beta x \right )^2} \, \mathrm{d}x \overset{x \mapsto 1/x}{=\! =\! =\! =\! =\! =\!} - \int_{0}^{\infty} \frac{\ln x}{\left ( \alpha x + \beta \right )^2} \, \mathrm{d}x}$
Οπότε,

$\displaystyle{\begin{aligned} \mathcal{J} &= \frac{1}{\alpha^2 - \beta^2} \int_{0}^{\infty} \frac{\ln x}{\left ( \alpha x + \beta \right )^2} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{\alpha x \mapsto \beta t}{=\! =\! =\! =\! =\!} \frac{1}{\alpha \beta \left ( \alpha^2 - \beta^2 \right )} \int_{0}^{\infty} \frac{\ln \frac{\beta}{\alpha}t}{\left ( 1+t \right )^2}\, \mathrm{d}t \\ &\!\!\!\! \!\overset{t \mapsto 1/t}{=\! =\! =\! =\! =\!} \frac{1}{\alpha \beta \left ( \alpha^2 - \beta^2 \right )} \int_{0}^{\infty} \frac{\ln \frac{\beta}{\alpha} \frac{1}{t}}{\left ( 1+t^2 \right )} \, \mathrm{d}t \\ &= \frac{1}{2 \alpha \beta \left ( \alpha^2 - \beta^2 \right )} \left ( \int_{0}^{\infty} \frac{\ln \frac{\beta}{\alpha}t}{\left ( 1+t \right )^2}\, \mathrm{d}t + \int_{0}^{\infty} \frac{\ln \frac{\beta}{\alpha} \frac{1}{t}}{\left ( 1+t \right )^2} \, \mathrm{d}t \right ) \\ &= \frac{1}{2 \alpha \beta \left ( \alpha^2 - \beta^2 \right )} \int_{0}^{\infty} \frac{2 \ln \frac{\beta}{\alpha}}{\left ( 1+t \right )^2} \, \mathrm{d}t \\ &= \frac{\ln \frac{\beta}{\alpha}}{\alpha \beta \left ( \alpha^2 - \beta^2 \right )} \cancelto{1}{\int_{0}^{\infty} \frac{\mathrm{d}t}{\left ( 1+t \right )^2}} \\ & = \frac{1}{\alpha \beta \left ( \alpha^2 - \beta^2 \right )} \ln \frac{\beta}{\alpha} \end{aligned}}$

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