# Show that $\int^\infty_1\frac{1}{1+x^3}dx = \int^1_0\frac{x}{1+x^3}dx$

Consider the integral $$J = \int^\infty_0\frac{1}{1+x^3}dx$$show that $$\int^\infty_1\frac{1}{1+x^3}dx = \int^1_0\frac{x}{1+x^3}dx$$and then deduce that $$J = \int^1_0f(x) dx$$ where f is a function to be determined.

I'm specifically stuck on the second part of the question. It is easy to miss but the bounds for J is $$0$$ and $$\infty$$ and not $$1$$ and $$\infty$$ as in the case of the first part of the question.

• Try setting $x=1/y$ May 7, 2019 at 8:59

Substitute $$u=1/x$$, that will solve it. A substition like this is often suitable, since the bounds in the first integral are $$1$$ and $$\infty$$ and the 'reciprocals' of these are $$1$$ and $$0$$, respectively.

Let ,

$$x = \frac{1}{t}$$ , $$dx = \frac{-1}{t^2}dt$$

at $$x = \infty, t = 0$$

at $$x = 1, t= 1$$

$$I = \int^0_1\frac{1}{1+\frac{1}{t^3}}.\frac{-dt}{t^2} = - \int^0_1\frac{t^3dt}{(t^3 + 1)t^2}$$

Changing the limits,

$$I = \int^1_0 \frac{tdt}{1+t^3} = \int^1_0\frac{xdx}{1+x^3}$$ (Replacing t by x)

$$J = \int^\infty_0\frac{dx}{1+x^3} = \int^1_0\frac{dx}{1+x^3}+ \int^\infty_1\frac{dx}{1+x^3} = \int^1_0\frac{dx}{1+x^3} + \int^1_0\frac{xdx}{1+x^3}$$ (From I)

$$J = \int^1_0\frac{(x+1)dx}{1+x^3}$$

Thus, $$f(x) = \frac{x+1}{1+x^3}$$

• Thanks. What about the deduce part? May 7, 2019 at 10:24
• I think $f(x) = \frac{x}{1+x^3}$, which is the integrand . May 7, 2019 at 10:28
• The bounds for J are 0 and $\infty$ not 1 and $\infty$. It is easy to miss. May 7, 2019 at 10:31
• Yes I've edited it now. May 7, 2019 at 10:40