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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.

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    $\begingroup$ Try setting $x=1/y$ $\endgroup$ – lab bhattacharjee May 7 '19 at 8:59
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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, $f(x) = \frac{x+1}{1+x^3}$

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  • $\begingroup$ Thanks. What about the deduce part? $\endgroup$ – Tightrope May 7 '19 at 10:24
  • $\begingroup$ I think $f(x) = \frac{x}{1+x^3} $, which is the integrand . $\endgroup$ – Ak. May 7 '19 at 10:28
  • $\begingroup$ The bounds for J are 0 and $\infty$ not 1 and $\infty$. It is easy to miss. $\endgroup$ – Tightrope May 7 '19 at 10:31
  • $\begingroup$ Yes I've edited it now. $\endgroup$ – Ak. May 7 '19 at 10:40
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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.

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