Menu Close

Find-0-1-E-x-E-x-dx-

Tinku Tara 0 Comments
Pin




Question Number 216749 by sniper237 last updated on 17/Feb/25
Find ∫_0 ^∞ (((−1)^(E(x)) )/(E(−x)))dx
Find∫0∞(−1)E(x)E(−x)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
∫0∞(−1)E(x)E(−x)=∫10(−1)0e(−x)+∫12(−1)1e(−x)+∫23(−1)2e(−x)+…=−(∫01dx−∫12dx2+∫23dx3−∫34dx4−…)=−(1−12+13−14−…)=−ln2
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=∑n=0∞∫nn+1(−1)n[−x]dxn⩽x<n+1⇒−n−1<−x<−n⇒[−x]=−n−1⇒I=∑n=0∞∫nn+1(−1)n−n−1dx=−∑n=0∞(−1)nn+1=−∑n=1∞(−1)n−1n=∑n=1∞(−1)nn=−ln2

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *