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Question: $\sum_{n=0}^{\infty} S_n$ is equal to...

∑n=0∞Sn\sum_{n=0}^{\infty} S_n is equal to

A

1+eπ1+e−π\frac{1+e^{\pi}}{1+e^{-\pi}}

B

12(1+e−π)(1−e−π)\frac{1}{2} \frac{(1+e^{-\pi})}{(1-e^{-\pi})}

C

12(1−e−π)\frac{1}{2(1-e^{-\pi})}

D

none of these

Answer

12(1+e−π)(1−e−π)\frac{1}{2} \frac{(1+e^{-\pi})}{(1-e^{-\pi})}

Explanation

Solution

The area SnS_n is given by the integral of the absolute value of the function y=e−xsin⁡xy = e^{-x} \sin x from x=nπx = n\pi to x=(n+1)πx = (n+1)\pi.

Sn=∫nπ(n+1)π∣e−xsin⁡x∣dxS_n = \int_{n\pi}^{(n+1)\pi} |e^{-x} \sin x| dx.

Since e−x>0e^{-x} > 0 for all xx, this is Sn=∫nπ(n+1)πe−x∣sin⁡x∣dxS_n = \int_{n\pi}^{(n+1)\pi} e^{-x} |\sin x| dx.

In the interval [nπ,(n+1)π][n\pi, (n+1)\pi], the sign of sin⁡x\sin x is constant, given by (−1)n(-1)^n.

So, ∣sin⁡x∣=∣(−1)nsin⁡(x−nπ)∣=∣sin⁡(x−nπ)∣|\sin x| = |(-1)^n \sin(x-n\pi)| = |\sin(x-n\pi)|. For x∈[nπ,(n+1)π]x \in [n\pi, (n+1)\pi], let u=x−nπu = x-n\pi, then u∈[0,π]u \in [0, \pi]. In this interval, sin⁡u≥0\sin u \geq 0, so ∣sin⁡u∣=sin⁡u|\sin u| = \sin u.

Thus, ∣sin⁡x∣=sin⁡(x−nπ)|\sin x| = \sin(x-n\pi).

Let's use the substitution u=x−nπu = x-n\pi in the integral for SnS_n. When x=nπx=n\pi, u=0u=0. When x=(n+1)πx=(n+1)\pi, u=πu=\pi. dx=dudx = du.

x=u+nπx = u+n\pi, so e−x=e−(u+nπ)=e−ue−nπe^{-x} = e^{-(u+n\pi)} = e^{-u} e^{-n\pi}.

∣sin⁡x∣=sin⁡(x−nπ)=sin⁡u|\sin x| = \sin(x-n\pi) = \sin u.

Sn=∫0πe−ue−nπsin⁡udu=e−nπ∫0πe−usin⁡uduS_n = \int_0^{\pi} e^{-u} e^{-n\pi} \sin u du = e^{-n\pi} \int_0^{\pi} e^{-u} \sin u du.

Let A=∫0πe−usin⁡uduA = \int_0^{\pi} e^{-u} \sin u du. We can evaluate this integral using integration by parts.

Let I=∫e−usin⁡uduI = \int e^{-u} \sin u du.

Using integration by parts ∫fg′=fg−∫f′g\int f g' = f g - \int f' g:

Let f=sin⁡uf = \sin u, g′=e−ug' = e^{-u}. Then f′=cos⁡uf' = \cos u, g=−e−ug = -e^{-u}.

I=−e−usin⁡u−∫cos⁡u(−e−u)du=−e−usin⁡u+∫e−ucos⁡uduI = -e^{-u} \sin u - \int \cos u (-e^{-u}) du = -e^{-u} \sin u + \int e^{-u} \cos u du.

Now evaluate ∫e−ucos⁡udu\int e^{-u} \cos u du. Let f=cos⁡uf = \cos u, g′=e−ug' = e^{-u}. Then f′=−sin⁡uf' = -\sin u, g=−e−ug = -e^{-u}.

∫e−ucos⁡udu=−e−ucos⁡u−∫(−sin⁡u)(−e−u)du=−e−ucos⁡u−∫e−usin⁡udu=−e−ucos⁡u−I\int e^{-u} \cos u du = -e^{-u} \cos u - \int (-\sin u) (-e^{-u}) du = -e^{-u} \cos u - \int e^{-u} \sin u du = -e^{-u} \cos u - I.

Substituting this back into the expression for II:

I=−e−usin⁡u+(−e−ucos⁡u−I)I = -e^{-u} \sin u + (-e^{-u} \cos u - I)

2I=−e−u(sin⁡u+cos⁡u)2I = -e^{-u} (\sin u + \cos u)

I=−12e−u(sin⁡u+cos⁡u)+CI = -\frac{1}{2} e^{-u} (\sin u + \cos u) + C.

Now evaluate the definite integral AA:

A=[−12e−u(sin⁡u+cos⁡u)]0πA = [-\frac{1}{2} e^{-u} (\sin u + \cos u)]_0^{\pi}

A=(−12e−π(sin⁡π+cos⁡π))−(−12e−0(sin⁡0+cos⁡0))A = \left(-\frac{1}{2} e^{-\pi} (\sin \pi + \cos \pi)\right) - \left(-\frac{1}{2} e^{-0} (\sin 0 + \cos 0)\right)

A=(−12e−π(0−1))−(−12⋅1⋅(0+1))A = \left(-\frac{1}{2} e^{-\pi} (0 - 1)\right) - \left(-\frac{1}{2} \cdot 1 \cdot (0 + 1)\right)

A=12e−π+12=12(1+e−π)A = \frac{1}{2} e^{-\pi} + \frac{1}{2} = \frac{1}{2} (1 + e^{-\pi}).

So, Sn=e−nπA=e−nπ12(1+e−π)S_n = e^{-n\pi} A = e^{-n\pi} \frac{1}{2} (1 + e^{-\pi}).

We need to find the sum ∑n=0∞Sn\sum_{n=0}^{\infty} S_n.

∑n=0∞Sn=∑n=0∞e−nπ12(1+e−π)\sum_{n=0}^{\infty} S_n = \sum_{n=0}^{\infty} e^{-n\pi} \frac{1}{2} (1 + e^{-\pi})

∑n=0∞Sn=12(1+e−π)∑n=0∞(e−π)n\sum_{n=0}^{\infty} S_n = \frac{1}{2} (1 + e^{-\pi}) \sum_{n=0}^{\infty} (e^{-\pi})^n.

This is an infinite geometric series with first term a=(e−π)0=1a = (e^{-\pi})^0 = 1 and common ratio r=e−πr = e^{-\pi}.

Since π>0\pi > 0, e−π<1e^{-\pi} < 1, so the series converges.

The sum of the geometric series is a1−r=11−e−π\frac{a}{1-r} = \frac{1}{1 - e^{-\pi}}.

Therefore, ∑n=0∞Sn=12(1+e−π)11−e−π=121+e−π1−e−π\sum_{n=0}^{\infty} S_n = \frac{1}{2} (1 + e^{-\pi}) \frac{1}{1 - e^{-\pi}} = \frac{1}{2} \frac{1 + e^{-\pi}}{1 - e^{-\pi}}.