How to prove ∫ 0 1 1 + x 30 1 + x 60 d x = 1 + c 31 , where 0

Question

How to prove       ∫    0    1              1      +              x                  30                            1      +              x                  60                      d  x  =  1  +      c    31  , where 0

Cecelia Mullins · Accepted Answer

we can use the power series expansion of

\frac{1}{1 + x^{60}}

to conclude that for

- 1 \leq x \leq 1

,

\frac{1 + x^{30}}{1 + x^{60}} = (1 + x^{30}) (1 - x^{60} + x^{120} - x^{180} + \dots) .

Multiply out, and integrate term by term. We get that our integral I is given by

I = 1 + \frac{1}{31} - \frac{1}{61} - \frac{1}{91} + \frac{1}{121} + \dots .

Group terms by twos, either starting at the beginning, or starting after the first term. From the two groupings, we can see that

1 < I < 1 + \frac{1}{31} .

We can use the same idea to get a string of increasingly more precise inequalities, such as

1 + \frac{1}{31} - \frac{1}{61} - \frac{1}{91} < I < 1 + \frac{1}{31} - \frac{1}{61} .

How to prove <msubsup> ∫ 0 1 </msubsup> 1

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