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Question Number 50659 by rahul 19 last updated on 18/Dec/18

$${\int }_0^{\mathrm{\infty }}\frac{dx}{4-x^2}=?$$

Commented by maxmathsup by imad last updated on 18/Dec/18

$$thisintegraldivergesletprovethislet{A}_n={\int }_0^{2+\frac{1}{n}}\frac{dx}{4-x^2}$$$$wehave{\int }_0^2\frac{dx}{4-x^2}={lim}_{n\to +\mathrm{\infty }}{A}_{n}but$$$$A_n=\frac{1}{4}{\int }_0^{2+\frac{1}{n}}(\frac{1}{2-x}+\frac{1}{2+x})dx=\frac{1}{4}{[ln\mid \frac{2+x}{2-x}\mid ]}_0^{2+\frac{1}{n}}$$$$=\frac{1}{4}\{ln\mid \frac{4+\frac{1}{n}}{-\frac{1}{n}}\mid \}=\frac{1}{4}ln(4n+1)\Rightarrow {lim}_{n\to +\mathrm{\infty }}{A}_n=+\mathrm{\infty }$$

Commented by mr W last updated on 18/Dec/18

$$sowecannotsay$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{x}dx=0?$$

Commented by maxmathsup by imad last updated on 18/Dec/18

$$thisintegralisanothertypebecausex\to \frac{1}{x}isodd.$$

Commented by maxmathsup by imad last updated on 18/Dec/18

$$sirwegetaindeterminedformbecause$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x}={\int }_{-\mathrm{\infty }}^0\frac{dx}{x}+{\int }_0^{+\mathrm{\infty }}\frac{dx}{x}and{\int }_{-\mathrm{\infty }}^0\frac{dx}{x}{=}_{x=-t}-{\int }_0^{+\mathrm{\infty }}\frac{-dt}{-t}=-{\int }_0^{\mathrm{\infty }}\frac{dx}{x}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x}={\int }_0^{\mathrm{\infty }}\frac{dx}{x}-{\int }_0^{\mathrm{\infty }}\frac{dx}{x}=+\mathrm{\infty }-\mathrm{\infty }indeterminedform!$$

Answered by mr W last updated on 18/Dec/18

$${\int }_0^{\mathrm{\infty }}\frac{dx}{2^2-x^2}$$$$=\frac{1}{4}{[\ln \mid \frac{2+x}{2-x}\mid ]}_0^2+\frac{1}{4}{[\ln \mid \frac{2+x}{2-x}\mid ]}_2^{\mathrm{\infty }}$$$$=\frac{1}{4}\left(A\right)+\frac{1}{4}(-A)$$$$=0$$

Commented by rahul 19 last updated on 18/Dec/18

thanks sir!

Commented by mr W last updated on 18/Dec/18

$$iamnotsureiftheansweriscorrect,$$$$sinceA\to \mathrm{\infty }.wecannotsay\mathrm{\infty }-\mathrm{\infty }=0.$$

Commented by afachri last updated on 18/Dec/18

$$\mathrm{that}\mathrm{is}\mathrm{the}\mathrm{thing}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{thinking}\mathrm{about}\mathrm{Mr}\mathrm{W}.$$$$\mathrm{i}\mathrm{was}\mathrm{going}\mathrm{to}\mathrm{ask}\mathrm{you},\mathrm{but}\mathrm{i}\mathrm{was}\mathrm{afraid}\mathrm{you}$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{want}\mathrm{to}\mathrm{answer}:\mathrm{D}$$

Answered by MJS last updated on 18/Dec/18

$$\int \frac{dx}{4-x^2}=\frac{1}{4}(\ln \mid x+2\mid -\ln \mid x-2\mid )+C$$$$\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}x=\pm 2$$$$\mathrm{Cauchy}\mathrm{Principal}\mathrm{Value}:$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{4-x^2}=\underset{\delta \to 0}{\lim }(\underset{0}{\stackrel{2-\delta }{\int }}\frac{dx}{4-x^2}+\underset{2+\delta }{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{4-x^2})=$$$$=\underset{\delta \to 0}{\lim }(\frac{1}{4}((\ln \mid 4-\delta \mid -\ln \mid -\delta \mid )-(\ln \mid 2\mid -\ln \mid -2\mid )+$$$$+(\underset{x\to \mathrm{\infty }}{\lim }(\ln \mid x+2\mid -\ln \mid x-2\mid )-(\ln \mid 4+\delta \mid -\ln \mid \delta \mid )=0$$$$\left(\mathrm{as}\mathrm{one}\mathrm{can}\mathrm{easily}\mathrm{see}\right)$$

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