Question Number 45735 by last updated on 16/Oct/18
$$\int_{\alpha} ^{\beta} \frac{\mathrm{1}}{\left({x}−\alpha\right)\left(\beta−{x}\right)}{dx}\:\:=?\:\:\:\beta>\alpha \\ $$
Commented by maxmathsup by imad last updated on 17/Oct/18
![I =−∫_α ^β (dx/((x−α)(x−β))) =−(1/(α−β))∫_α ^β { (1/(x−α)) −(1/(x−β))} =(1/(β−α)) [ln∣((x−α)/(x−β))∣]_α ^β =(1/(β−α)){ln(β−α)−ln(0^+ )−ln(0^+ )+ln(β−α)} =(1/(β−α)){ 2ln(β−α) +∞} =+∞ and this integral diverves to+∞=!.](Q45886.png)
$${I}\:=−\int_{\alpha} ^{\beta} \:\:\:\:\:\frac{{dx}}{\left({x}−\alpha\right)\left({x}−\beta\right)}\:=−\frac{\mathrm{1}}{\alpha−\beta}\int_{\alpha} ^{\beta} \left\{\:\frac{\mathrm{1}}{{x}−\alpha}\:−\frac{\mathrm{1}}{{x}−\beta}\right\} \\ $$$$=\frac{\mathrm{1}}{\beta−\alpha}\:\left[{ln}\mid\frac{{x}−\alpha}{{x}−\beta}\mid\right]_{\alpha} ^{\beta} =\frac{\mathrm{1}}{\beta−\alpha}\left\{{ln}\left(\beta−\alpha\right)−{ln}\left(\mathrm{0}^{+} \right)−{ln}\left(\mathrm{0}^{+} \right)+{ln}\left(\beta−\alpha\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\beta−\alpha}\left\{\:\mathrm{2}{ln}\left(\beta−\alpha\right)\:+\infty\right\}\:=+\infty\:\:{and}\:{this}\:{integral}\:{diverves}\:{to}+\infty=!. \\ $$
Answered by ARVIND DADHICH last updated on 16/Oct/18
![(1/(β−α))∫_α ^β ((β−x+x−α)/((x−α)(β−x)))dx (1/(β−α))∫_α ^β (1/(x−α))+(1/(β−x))dx (1/(β−α))[ln(x−α)+ln(β−x)]_α ^β (1/(β−α))[ln(β−α)−ln(β−α)] =0](Q45736.png)
$$\frac{\mathrm{1}}{\beta−\alpha}\int_{\alpha} ^{\beta} \frac{\beta−{x}+{x}−\alpha}{\left({x}−\alpha\right)\left(\beta−{x}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\beta−\alpha}\int_{\alpha} ^{\beta} \frac{\mathrm{1}}{{x}−\alpha}+\frac{\mathrm{1}}{\beta−{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\beta−\alpha}\left[{ln}\left({x}−\alpha\right)+{ln}\left(\beta−{x}\right)\right]_{\alpha} ^{\beta} \\ $$$$\frac{\mathrm{1}}{\beta−\alpha}\left[{ln}\left(\beta−\alpha\right)−{ln}\left(\beta−\alpha\right)\right] \\ $$$$=\mathrm{0} \\ $$
Commented by last updated on 16/Oct/18
$${sir}\:{ln}\mathrm{0}=\mathrm{0}\:{this}\:{is}\:{right} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18
$${ln}\left(\mathrm{0}\right)=−\infty\:\:\:\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18
$${for}\:{both}\:\alpha\:{and}\:\beta\:{denominator}\:{is}\:{zero} \\ $$$${so}\:{it}\:{is}\:{improper}\:{intregal}... \\ $$$$=\underset{{h}\rightarrow\alpha} {\mathrm{li}\overset{\underset{{k}\rightarrow\beta} {\mathrm{lim}}} {\mathrm{m}}}\int_{{h}} ^{{k}} \frac{{dx}}{\left({x}−\alpha\right)\left(\beta−{x}\right)} \\ $$$$ \\ $$
Commented by last updated on 17/Oct/18
$${thanks}\:{sir} \\ $$
Answered by MJS last updated on 16/Oct/18
$$\int\frac{{dx}}{\left({x}−\alpha\right)\left(\beta−{x}\right)}=−\int\frac{{dx}}{\left({x}−\alpha\right)\left({x}−\beta\right)}= \\ $$$$=\frac{\mathrm{1}}{\beta−\alpha}\int\left(\frac{\mathrm{1}}{{x}−\alpha}−\frac{\mathrm{1}}{{x}−\beta}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{\beta−\alpha}\left(\mathrm{ln}\:\mid{x}−\alpha\mid\:−\mathrm{ln}\:\mid{x}−\beta\mid\right)={F}\left({x}\right) \\ $$$${F}\left(\beta−\delta\right)−{F}\left(\alpha+\delta\right)=\frac{\mathrm{1}}{\beta−\alpha}\left(\mathrm{ln}\:\mid\beta−\alpha−\delta\mid\:−\mathrm{ln}\:\mid\delta\mid\:−\mathrm{ln}\:\mid\delta\mid\:+\mathrm{ln}\:\mid\alpha−\beta+\delta\mid\right)= \\ $$$$=\frac{\mathrm{2}}{\beta−\alpha}\left(\mathrm{ln}\:\mid\alpha−\beta+\delta\mid\:−\mathrm{ln}\:\mid\delta\mid\right) \\ $$$$\underset{\delta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}}{\beta−\alpha}\left(\mathrm{ln}\:\mid\alpha−\beta+\delta\mid\:−\mathrm{ln}\:\mid\delta\mid\right)=+\infty \\ $$$$ \\ $$