Proving $\int_0^\infty \frac{x}{(1+x)^3} dx = \frac{1}{2} \int_0^\infty \frac{1}{(1+x)^2} dx$

Question

To prove: $$\int_0^\infty \frac{x}{(1+x)^3} dx = \frac{1}{2} \int_0^\infty \frac{1}{(1+x)^2} dx$$

I don't know how to integrate these integrals. I tried integration by parts and it doesn't get me through. How can i prove this?

Answer 1

$\int_0^{\infty} \frac{x}{(1+x)^3}dx=\int_0^{\infty} x\ast \frac{1}{(1+x)^3}dx=\int_0^{\infty} u\ast v^{'}=u\ast v\Bigg|_0^{\infty} -\int_0^\infty u^{'}\ast v = x\ast (-\frac{1}{2(1+x)^2})\Bigg|_0^{\infty} -\int_0^\infty1\ast (-\frac{1}{2(1+x)^2})dx=0-\int_0^{\infty} -\frac{1}{2(1+x)^2}dx= \\ =\frac{1}{2}\int_0^{\infty} \frac{1}{(1+x)^2}dx \blacksquare$

Answer 2

$$\begin{array}{l}\displaystyle\int\limits_0^\infty {\frac{x}{{{{\left( {1 + x} \right)}^3}}}dx} ,\left\{ \begin{array}{l}u = x\\dv = \frac{{dx}}{{{{\left( {1 + x} \right)}^3}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = - \frac{1}{2}\frac{1}{{{{\left( {1 + x} \right)}^2}}}\end{array} \right.\\ \displaystyle\Rightarrow \int\limits_0^\infty {\frac{x}{{{{\left( {1 + x} \right)}^3}}}dx} = \left. { - \frac{1}{2}\frac{x}{{{{\left( {1 + x} \right)}^2}}}} \right|_0^\infty + \frac{1}{2}\int\limits_0^\infty {\frac{1}{{{{\left( {1 + x} \right)}^2}}}dx} \\\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{\left( {1 + x} \right)}^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{x{{\left( {1 + \frac{1}{x}} \right)}^2}}} = 0 \Rightarrow \int\limits_0^\infty {\frac{x}{{{{\left( {1 + x} \right)}^3}}}dx} = \frac{1}{2}\int\limits_0^\infty {\frac{1}{{{{\left( {1 + x} \right)}^2}}}dx} \end{array}$$

Answer 3

This kind of integration is quite easy if you know Beta Gamma functions

In one of its forms $$ \beta(m,n)=\int_{0}^{\infty} \frac{t^{m-1}}{(1+t)^{m+n}} dt $$

So the given integrals are $\beta(2,1)$ and $\beta(1,1)$ $$\beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$

For an integer m

$\Gamma(m)=(m-1)!$

So the question simplifies to $$\frac{1!0!}{2!} = \frac{1}{2}\frac{0!0!}{1!}$$

There is good further reading in NP.Bali Real analysis if u want to check it out

You could also use by parts by taking limits but its basically reinventing the wheel

Answer 4

$$\frac 12 \int_0^{\infty} \frac{1}{(1+x)^2}dx$$ Considering $u=\frac{1}{(1+x)^2}$, $v=1$,

$$\frac 12 \int_0^{\infty} \frac{1}{(1+x)^2}dx$$ $$=\frac 12 \left[\frac{x}{(1+x)^2}+2\int\frac{x}{(1+x)^3}dx\right]_0^\infty$$ $$=\frac 12 \lim_{x \to \infty}\frac{x}{(1+x)^2}+\int_0^\infty\frac{x}{(1+x)^3}dx$$ Using LHospitals rule on the limits problem, we get it to be $0$. Hence, $$\frac 12 \int_0^{\infty} \frac{1}{(1+x)^2}dx=\int_0^\infty\frac{x}{(1+x)^3}dx$$

Answer 5

$$ \begin{aligned} \int_0^{\infty} \frac{x}{(1+x)^3} d x & =-\frac{1}{2} \int_0^{\infty} x \,d \left[\frac{1}{(1+x)^2}\right] \\ & =-\left[\frac{x}{2(1+x)^2}\right]_0^{\infty}+\frac{1}{2} \int_0^{\infty} \frac{d x}{(1+x)^2} \\ & =\frac{1}{2} \int_0^{\infty} \frac{d x}{(1+x)^2} \end{aligned} $$

Answer 6

In fact, let $$I=2\int_0^\infty \frac{x}{(1+x)^3} dx - \int_0^\infty \frac{1}{(1+x)^2} dx=\int_0^\infty\frac{x-1}{(1+x)^3}dx. $$ Now $$I\overset{x\to\frac1x}=\int_0^\infty\frac{\frac1x-1}{(1+\frac1x)^3}\frac{dx}{x^2}=\int_0^\infty\frac{1-x}{(1+x)^3}dx=-I$$ and hence $I=0$.