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Evaluate: $$\int_0^{\infty } {\frac{(x^7)(1-x^{12})}{(1+x)^{28}}}dx$$

The answer is zero, but I cannot seem to figure out the steps.

Eleven-Eleven
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Ally
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  • FYI:http://www.wolframalpha.com/input/?i=integral_0%5E%28infinity%29+%28%28x%5E%287%29-x%5E%2819%29%29%2F%281%2Bx%29%5E28%29dx – mathlove Dec 19 '13 at 10:40

1 Answers1

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Let us denote the integral by $I$. Applying the substitution $x \mapsto 1/x$, we have

$$ I = \int_{0}^{\infty} \frac{x^{-7}(1 - x^{-12})}{(1 + x^{-1})^{28}} \, \frac{dx}{x^{2}} = \int_{0}^{\infty} \frac{x^{7}(x^{12} - 1)}{(1 + x)^{28}} \, dx = -I. $$

Therefore $I = 0$.

Sangchul Lee
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  • Cool, and relies on the fact that $2 + 7 + 12 = 28 - 7$, meaning the result is not zero unless this substitution results in $-I$. – Bennett Gardiner Dec 19 '13 at 12:44
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    @BennettGardiner, Exactly. The numbers are deliberately chosen to satsify that relations. In general, $$ \int_{0}^{\infty} \frac{x^{a}(1-x^{b})}{(1+x)^{c}} , dx = \frac{\Gamma(a+1)\Gamma(c-a-1)-\Gamma(a+b+1)\Gamma(c-a-b-1)}{\Gamma(c)}.$$ – Sangchul Lee Dec 19 '13 at 12:53
  • So what is the equation we get when we set the RHS to zero? I assume the Gamma functions simplify to an algebraic (linear?) relationship? – Bennett Gardiner Dec 20 '13 at 00:33