Integral Sequence

Calculus Level 3

∫ 0 i 2 π e x 2 ( 2 − e x ) d x + ∫ 0 i 2 π e 2 x 4 ( 2 − e x ) d x + ∫ 0 i 2 π e 3 x 8 ( 2 − e x ) d x + ⋯ = ? \int_{0}^{i2\pi} \frac{e^{x}}{2(2-e^{x})} \, dx + \int_{0}^{i2\pi} \frac{e^{2x}}{4(2-e^{x})}\, dx+ \int_{0}^{i2\pi} \frac{e^{3x}}{8(2-e^{x})}\, dx +\cdots = \, ?

Clarification : i = − 1 i=\sqrt{-1} .


The answer is 0.

Mateo Reddy by Mateo Reddy , 21, Switzerland

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1 solution

Mateo Reddy
Jun 26, 2016

The sequence can be rewritten as

∫ 0 i 2 π 1 2 e x + 1 4 e 2 x + 1 8 e 3 x + . . . 2 − e x d x \int_{0}^{i2\pi}\frac{\frac{1}{2}e^{x}+\frac{1}{4}e^{2x}+\frac{1}{8}e^{3x}+...}{2-e^{x}}dx

common ratio = 1 2 e x =\frac{1}{2}e^{x}

S ∞ = 1 2 e x 1 − 1 2 e x = e x 2 − e x S_{\infty}=\frac{\frac{1}{2}e^{x}}{1-\frac{1}{2}e^{x}}=\frac{e^{x}}{2-e^{x}}

Now we have the following integral

∫ 0 i 2 π e x 2 − e x 2 − e x d x = ∫ 0 i 2 π e x ( 2 − e x ) 2 d x \int_{0}^{i2\pi}\frac{\frac{e^{x}}{2-e^{x}}}{2-e^{x}}dx=\int_{0}^{i2\pi}\frac{e^{x}}{(2-e^{x})^{2}}dx

Using u-substitution

u = 2 − e x u=2-e^{x}

d u d x = − e x \frac{du}{dx}=-e^{x}

d x = − 1 e x d u dx=-\frac{1}{e^{x}}du

∫ 0 i 2 π e x u 2 × ( − 1 e x ) d u = − ∫ 0 i 2 π 1 u 2 d u = [ 1 u ] 0 i 2 π \int_{0}^{i2\pi}\frac{e^{x}}{u^{2}}\times \left ( -\frac{1}{e^{x}} \right )du=-\int_{0}^{i2\pi}\frac{1}{u^{2}}du=\left [ \frac{1}{u} \right ]_{0}^{i2\pi}

but u = 2 − e x u=2-e^{x}

∴ [ 1 2 − e x ] 0 i 2 π = 1 2 − e i 2 π − 1 2 − e 0 = 1 − 1 = 0 \therefore \left [ \frac{1}{2-e^{x}} \right ]_{0}^{i2\pi}=\frac{1}{2-e^{i2\pi}}-\frac{1}{2-e^{0}}=1-1=\mathbf{0}

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