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Question Number 216749 by sniper237 last updated on 17/Feb/25

Find ∫_0 ^∞ (((−1)^(E(x)) )/(E(−x)))dx

$${Find}\:\:\int_{\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{E}\left({x}\right)} }{{E}\left(−{x}\right)}{dx} \\ $$

Answered by mehdee7396 last updated on 18/Feb/25

∫_0 ^∞ (((−1)^(E(x)) )/(E(−x)))=∫_0 ^1 (((−1)^0 )/(e(−x)))+∫_1 ^2 (((−1)^1 )/(e(−x)))+∫_2 ^3 (((−1)^2 )/(e(−x)))+... =−(∫_0 ^1 dx−∫_1 ^2 (dx/2)+∫_2 ^3 (dx/3)−∫_3 ^4 (dx/4)−...) =−(1−(1/2)+(1/3)−(1/4)−...) =−ln2

$$\int_{\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{E}\left({x}\right)} }{{E}\left(−{x}\right)}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left(−\mathrm{1}\right)^{\mathrm{0}} }{{e}\left(−{x}\right)}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left(−\mathrm{1}\right)^{\mathrm{1}} }{{e}\left(−{x}\right)}+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{{e}\left(−{x}\right)}+... \\ $$$$=−\left(\int_{\mathrm{0}} ^{\mathrm{1}} {dx}−\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dx}}{\mathrm{2}}+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{dx}}{\mathrm{3}}−\int_{\mathrm{3}} ^{\mathrm{4}} \frac{{dx}}{\mathrm{4}}−...\right) \\ $$$$=−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}−...\right) \\ $$$$=−{ln}\mathrm{2} \\ $$$$ \\ $$

Answered by mathmax last updated on 23/Feb/25

I =Σ_(n=0) ^∞ ∫_n ^(n+1) (((−1)^n )/([−x]))dx n≤x<n+1 ⇒−n−1<−x<−n ⇒ [−x]=−n−1 ⇒I=Σ_(n=0) ^∞ ∫_n ^(n+1) (((−1)^n )/(−n−1))dx =−Σ_(n=0) ^∞ (((−1)^n )/(n+1))=−Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =Σ_(n=1) ^∞ (((−1)^n )/n)=−ln2

$${I}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \int_{{n}} ^{{n}+\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left[−{x}\right]}{dx} \\ $$$${n}\leqslant{x}<{n}+\mathrm{1}\:\Rightarrow−{n}−\mathrm{1}<−{x}<−{n}\:\Rightarrow \\ $$$$\left[−{x}\right]=−{n}−\mathrm{1}\:\Rightarrow{I}=\sum_{{n}=\mathrm{0}} ^{\infty} \int_{{n}} ^{{n}+\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{−{n}−\mathrm{1}}{dx} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}=−{ln}\mathrm{2} \\ $$

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