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Let $$\int_0^{\infty}\frac{dx}{1+x^{2021}}=\int_0^1\frac{dx}{(1-x^{2021})^{a}}$$ then find $a$

This is one of the questions of the HOTS section of my practice sheet. I can not think of any way to solve it. I got some hint from limits of the integrals which made me substitute $x^\frac{2021}{2}=\tan t$ and $ x^\frac{2021}{2}=\sin t$ in first and second integral respectively which made both of their limits same. However this made the integrals more messier as I tried to substitute the expressions for $dx$ in both.

Any helps would be appreciated

Acc2
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1 Answers1

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Using your substitution

$$x^\frac{2021}{2}=\tan t\implies x={(\tan t)}^{\frac{2}{2021}}$$

The integral on LHS becomes

$$I_1=\frac{2}{2021}\int_{0}^{\pi/2}\frac{dt}{{(\tan t)}^{\frac{2019}{2021}}}$$

Using your substitution

$$x^\frac{2021}{2}=\sin t\implies x={(\sin t)}^{\frac{2}{2021}}$$

The integral on RHS becomes

$$I_2=\frac{2}{2021}\int_{0}^{\pi/2}\frac{(\cos t)^{1-2a}dt}{{(\sin t)}^{\frac{2019} {2021}}}$$

Compare both of them to get

$$\boxed{a=\frac{1}{2021}}$$

Lalit Tolani
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