Question Number 148000 by vvvv last updated on 25/Jul/21
$$\underset{−\infty} {\overset{+\infty} {\int}}\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$
Answered by gsk2684 last updated on 25/Jul/21
![2∫_(x=0) ^∞ (dx/(x^3 (1+(k^2 /x^2 ))^(3/2) )) changement 1+(k^2 /x^2 )=t ⇒ −((2k^2 )/x^3 )dx=dt 2∫_(t=∞) ^1 t^(−(3/2)) (dt/(−2k^2 ))=((−1)/k^2 )[(t^(((−3)/2)+1) /(−(3/2)+1))]_∞ ^1 =((−1)/k^2 )[(t^(−(1/2)) /(−(1/2)))]_∞ ^1 =(2/k^2 )[(1/( (√t)))]_∞ ^1 =(2/k^2 )[1−0]=(2/k^2 )](https://www.tinkutara.com/question/Q148002.png)
$$\mathrm{2}\underset{{x}=\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{{k}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\: \\ $$$${changement}\:\mathrm{1}+\frac{{k}^{\mathrm{2}} }{{x}^{\mathrm{2}} }={t}\:\Rightarrow\:−\frac{\mathrm{2}{k}^{\mathrm{2}} }{{x}^{\mathrm{3}} }{dx}={dt} \\ $$$$\mathrm{2}\underset{{t}=\infty} {\overset{\mathrm{1}} {\int}}{t}^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\frac{{dt}}{−\mathrm{2}{k}^{\mathrm{2}} }=\frac{−\mathrm{1}}{{k}^{\mathrm{2}} }\left[\frac{{t}^{\frac{−\mathrm{3}}{\mathrm{2}}+\mathrm{1}} }{−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}\right]_{\infty} ^{\mathrm{1}} \\ $$$$=\frac{−\mathrm{1}}{{k}^{\mathrm{2}} }\left[\frac{{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{−\frac{\mathrm{1}}{\mathrm{2}}}\right]_{\infty} ^{\mathrm{1}} =\frac{\mathrm{2}}{{k}^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\:\sqrt{{t}}}\right]_{\infty} ^{\mathrm{1}} =\frac{\mathrm{2}}{{k}^{\mathrm{2}} }\left[\mathrm{1}−\mathrm{0}\right]=\frac{\mathrm{2}}{{k}^{\mathrm{2}} } \\ $$
Answered by mathmax by abdo last updated on 25/Jul/21
![U_k =∫_(−∞) ^(+∞) (dx/((x^2 +k^2 )^(3/2) )) changement x=ktanθ give U_k =∫_(−(π/2)) ^(π/2) ((k(1+tan^2 θ))/(k^3 (1+tan^2 θ)^(3/2) ))dθ =(1/k^2 )∫_(−(π/2)) ^(π/2) (dθ/( (√(1+tan^2 θ)))) =(1/k^2 )∫_(−(π/2)) ^(π/2) cosθ dθ =(1/k^2 )[sinθ]_(−(π/2)) ^(π/2) =(1/k^2 )(1−(−1)) =(2/k^2 ) (k≠0)](https://www.tinkutara.com/question/Q148003.png)
$$\mathrm{U}_{\mathrm{k}} =\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:\mathrm{changement}\:\mathrm{x}=\mathrm{ktan}\theta\:\mathrm{give} \\ $$$$\mathrm{U}_{\mathrm{k}} =\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{k}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)}{\mathrm{k}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\mathrm{d}\theta\:=\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{d}\theta}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\theta\:\mathrm{d}\theta\:=\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\left[\mathrm{sin}\theta\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\left(\mathrm{1}−\left(−\mathrm{1}\right)\right)\:=\frac{\mathrm{2}}{\mathrm{k}^{\mathrm{2}} }\:\:\:\:\left(\mathrm{k}\neq\mathrm{0}\right) \\ $$