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I tried evaluating the integral $\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$ but I wasn't able to get the result. Following is the way by which I did it- $$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$ $$\implies I=\int_{0}^{\infty} x^{49}(1+x)^{-51}dx$$ Further, I tried Integration by parts but it didn't worked. Can anyone tell that how this integral can be evaluated.

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    $\begingroup$ HINT: try substitution $$u= \frac{x}{1+x}$$ $\endgroup$
    – Crostul
    Aug 18, 2021 at 3:21
  • $\begingroup$ sorry, @Mathzcreator, but larger titles are discouraged on math.SE so that your question does not take up more vertical space than others, so I've edited the title. See this discussion on the meta site: math.meta.stackexchange.com/questions/9687/… $\endgroup$ Aug 18, 2021 at 3:59
  • $\begingroup$ Note $$\int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx \overset{x\to\frac1x}= \int_{0}^{\infty} \dfrac{1}{(1+x)^{51}}dx =\frac1{50}$$ $\endgroup$
    – Quanto
    Feb 23 at 19:10

3 Answers 3

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The given integral is

$$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$

$$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{51}}{x^2(1+x)^{51}}dx$$

$$I=\displaystyle \int_{0}^{\infty} \dfrac{1}{x^2(1+\frac{1}{x})^{51}}dx$$

Let $u=1+\frac{1}{x}$ , therefore $\displaystyle du=-\frac{dx}{x^2}$

Therefore $$I=\displaystyle \int_\infty^1-\frac{du}{u^{51}}=\frac{1}{50}\bigg[\frac{1}{u^{50}}\bigg]^{1}_{\infty} = \boxed{\frac{1}{50}}$$

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$$I=\displaystyle\int_{0}^{\infty}\dfrac{x^{49}}{(1+x)^{51}}dx$$ $$I=\displaystyle\int_{0}^{\infty}\dfrac{x^{50-1}}{(1+x)^{50+1}}dx$$ Using Beta Function,

$$B(x,y)=\displaystyle\int_{0}^{\infty}\dfrac{t^{x-1}}{(1+t)^{x+y}}dt$$

$$\implies I=\displaystyle\int_{0}^{\infty}\dfrac{x^{50-1}}{(1+x)^{50+1}}dx=B(50,1)$$ Using Beta Function and Gamma Function Relationship,

$$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

$\implies I=B(50,1)=\dfrac{\Gamma(50)\Gamma(1)}{\Gamma(50+1)}=\dfrac{\Gamma(50)\Gamma(1)}{50\Gamma(50)}=\dfrac{\Gamma(1)}{50}=\dfrac{1}{50}$ $$\implies \boxed{I=\dfrac{1}{50}}$$

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Here is an alternative method.

First of all, for each integer $n>1$, we have $\int_1^\infty x^{-n}dx=\frac1{n-1}$.

Thus, using the binomial theorem, \begin{align} \int_0^\infty\frac{x^{49}}{(1+x)^{51}}dx&=\int_1^\infty\frac{(x-1)^{49}}{x^{51}}dx\\ &=\int_1^\infty\frac1{x^{51}}\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}x^ndx\\ &=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\int_1^\infty x^{n-51}dx\\ &=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\frac1{50-n}\\ &=\frac1{50}\sum_{n=0}^{49}(-1)^{n-1}{50\choose n}\\ &=-\frac1{50}\left((1+(-1))^{50}-{50\choose50}(-1)^{50}\right)\\ &=\frac1{50}. \end{align}

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