Integrate with power of e

Calculus Level 4

∫ − ∞ ∞ e − 6 x ( 1 + e − 2 x ) 7 2 d x = A B \large{\displaystyle \int^{\infty}_{-\infty} \frac{e^{-6x}}{(1+e^{-2x})^{\frac{7}{2}}}dx=\frac{A}{B}}

where A , B A,B are co prime positive integers.

Find B − A B-A


The answer is 7.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Chew-Seong Cheong
Sep 20, 2015

∫ − ∞ ∞ e − 6 x ( 1 + e − 2 x ) 7 2 d x = ∫ ∞ 1 ( u − 1 ) 3 − 2 ( u − 1 ) u 7 2 d x Let u = 1 + e − 2 x , e − 2 x = u − 1 , d u = − 2 e − 2 x d x = 1 2 ∫ 1 ∞ ( u − 1 ) 2 u 7 2 d u = 1 2 ∫ 1 ∞ u 2 − 2 u + 1 u 7 2 d u = 1 2 ∫ 1 ∞ ( 1 u 3 2 − 2 u 5 2 + 1 u 7 2 ) d u = 1 2 [ − 2 u 1 2 + 4 3 u 3 2 − 2 5 u 5 2 ] 1 ∞ = 1 2 [ 0 + 2 − 4 3 + 2 5 ] = 1 2 [ 30 − 20 + 6 15 ] = 8 15 \begin{aligned} \int_{-\infty}^{\infty} \frac{e^{-6x}}{(1+e^{-2x})^{\frac{7}{2}}} dx & = \int_{\infty}^{1} \frac{(u-1)^3}{-2(u-1)u^{\frac{7}{2}}} dx \quad \quad \small \color{#3D99F6}{\text{Let } u = 1+e^{-2x}, \space e^{-2x} = u-1, \space du = -2e^{-2x}dx} \\ & = \frac{1}{2} \int_{1}^{\infty} \frac{(u-1)^2}{u^{\frac{7}{2}}} du \\ & = \frac{1}{2} \int_{1}^{\infty} \frac{u^2-2u+1}{u^{\frac{7}{2}}} du \\ & = \frac{1}{2} \int_{1}^{\infty} \left( \frac{1}{u^{\frac{3}{2}}} - \frac{2}{u^{\frac{5}{2}}} + \frac{1}{u^{\frac{7}{2}}} \right) du \\ & = \frac{1}{2}\left[ - \frac{2}{u^{\frac{1}{2}}} + \frac{4}{3u^{\frac{3}{2}}} - \frac{2}{5u^{\frac{5}{2}}} \right]_1^\infty \\ & = \frac{1}{2}\left[ 0 + 2 - \frac{4}{3} + \frac{2}{5} \right] = \frac{1}{2}\left[ \frac{30-20+6}{15} \right] = \boxed{\dfrac{8}{15}} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nicely done sir. I did by putting e − x = tan ⁡ z e^{-x}=\tan z

Tanishq Varshney - 5 years, 8 months ago

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Ya best move to do it with beta-gamma function

Md Zuhair - 3 years, 11 months ago

I'm not seeing where you substituted dx to du. This is exactly how I did it, but your substitution is unclear since you leave dx when there needs to be a du.

M M - 5 years, 5 months ago

@Chew-Seong Cheong Sir , I think there are a few typos in the solution (dx should be replaced with du)

Ankit Kumar Jain - 3 years ago

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Yeah.. true. That I guess is a general reflex😂😂😂 to put DX after integral

Md Zuhair - 3 years ago

Thanks, I have changed them.

Chew-Seong Cheong - 3 years ago

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