Question Number 37898 by abdo mathsup 649 cc last updated on 19/Jun/18
![calculate f(λ) = ∫_0 ^(+∞) e^(−λx) cos(π[x])dx withλ>0](Q37898.png)
$${calculate}\:{f}\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−\lambda{x}} \:{cos}\left(\pi\left[{x}\right]\right){dx} \\ $$ $${with}\lambda>\mathrm{0} \\ $$
Commented byprof Abdo imad last updated on 19/Jun/18
![f(λ)=Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−λx) cos(nπ)dx =Σ_(n=0) ^∞ (−1)^n ∫_n ^(n+1) e^(−λx) dx =Σ_(n=0) ^∞ (−1)^n [−(1/λ) e^(−λx) ]_n ^(n+1) =(1/λ)Σ_(n=0) ^∞ (−1)^(n+1) ( e^(−λ(n+1)) −e^(−λn) ) =(1/λ)Σ_(n=0) ^∞ (−1)^n e^(−λn) −(1/λ)Σ_(n=0) ^∞ (−1)^n e^(−λ(n+1)) =(1/λ) Σ_(n=0) ^∞ (−e^(−λ) )^n −(e^(−λ) /λ) Σ_(n=0) ^∞ (−e^(−λ) )^n =((1−e^(−λ) )/λ) (1/(1+e^(−λ) )) ⇒ f(λ) = ((1−e^(−λ) )/(λ(1+e^(−λ) )))](Q37923.png)
$${f}\left(\lambda\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:{e}^{−\lambda{x}} \:{cos}\left({n}\pi\right){dx} \\ $$ $$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:{e}^{−\lambda{x}} {dx} \\ $$ $$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \left[−\frac{\mathrm{1}}{\lambda}\:{e}^{−\lambda{x}} \right]_{{n}} ^{{n}+\mathrm{1}} \\ $$ $$=\frac{\mathrm{1}}{\lambda}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:\left(\:{e}^{−\lambda\left({n}+\mathrm{1}\right)} \:−{e}^{−\lambda{n}} \right) \\ $$ $$=\frac{\mathrm{1}}{\lambda}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\lambda{n}} \:−\frac{\mathrm{1}}{\lambda}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\lambda\left({n}+\mathrm{1}\right)} \\ $$ $$=\frac{\mathrm{1}}{\lambda}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{e}^{−\lambda} \right)^{{n}} \:\:−\frac{{e}^{−\lambda} }{\lambda}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{e}^{−\lambda} \right)^{{n}} \\ $$ $$=\frac{\mathrm{1}−{e}^{−\lambda} }{\lambda}\:\:\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\lambda} }\:\Rightarrow \\ $$ $${f}\left(\lambda\right)\:=\:\frac{\mathrm{1}−{e}^{−\lambda} }{\lambda\left(\mathrm{1}+{e}^{−\lambda} \right)} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18
![[x]=0 1>x≥0 [x]=1 2>x≥1 [x]=2 3>x≥2 thus the value of Π[x] is intregal multiple of Π...hence value of cos(Π[x]) oscillates eitther +1 or −1 so the given intregal has two value 1)∫_0 ^(+∞) e^(−λx) ×1×dx =∣(e^(−λx) /(−λ))∣_0 ^∞ =((−1)/λ)((1/e^∞ )−(1/e^0 ))=(1/λ) 2)∫_0 ^(+∞) e^(−λx) ×−1×dx =−1×∣(e^(−λx) /(−λ))∣_0 ^∞ =(1/λ)((1/e^∞ )−(1/e^0 ))=−(1/λ)](Q37903.png)
$$\left[{x}\right]=\mathrm{0}\:\:\:\:\mathrm{1}>{x}\geqslant\mathrm{0} \\ $$ $$\left[{x}\right]=\mathrm{1}\:\:\:\mathrm{2}>{x}\geqslant\mathrm{1} \\ $$ $$\left[{x}\right]=\mathrm{2}\:\:\mathrm{3}>{x}\geqslant\mathrm{2} \\ $$ $$ \\ $$ $${thus}\:{the}\:{value}\:{of}\:\Pi\left[{x}\right]\:{is}\:{intregal}\:{multiple}\:{of} \\ $$ $$\Pi...{hence}\:{value}\:{of}\:{cos}\left(\Pi\left[{x}\right]\right)\:{oscillates} \\ $$ $${eitther}\:+\mathrm{1}\:{or}\:−\mathrm{1} \\ $$ $${so}\:{the}\:{given}\:{intregal}\:{has}\:{two}\:{value} \\ $$ $$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{+\infty} {e}^{−\lambda{x}} ×\mathrm{1}×{dx} \\ $$ $$=\mid\frac{{e}^{−\lambda{x}} }{−\lambda}\mid_{\mathrm{0}} ^{\infty} \\ $$ $$=\frac{−\mathrm{1}}{\lambda}\left(\frac{\mathrm{1}}{{e}^{\infty} }−\frac{\mathrm{1}}{{e}^{\mathrm{0}} }\right)=\frac{\mathrm{1}}{\lambda} \\ $$ $$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{+\infty} {e}^{−\lambda{x}} ×−\mathrm{1}×{dx} \\ $$ $$=−\mathrm{1}×\mid\frac{{e}^{−\lambda{x}} }{−\lambda}\mid_{\mathrm{0}} ^{\infty} \:=\frac{\mathrm{1}}{\lambda}\left(\frac{\mathrm{1}}{{e}^{\infty} }−\frac{\mathrm{1}}{{e}^{\mathrm{0}} }\right)=−\frac{\mathrm{1}}{\lambda} \\ $$ $$ \\ $$