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I have the following task:

First to prove that for $x \in [0,1)$ that $1-x\leq e^{-x} \leq \frac{1}{1+x}$

Then to deduce that $\int^1_0(1-x^2)^ndx\leq\int^1_0e^{-nx^2}dx\leq\int^1_0\frac{1}{(1+x^2)^n}dx$ for all $n \geq1$

So now, using the results above which i've completed, i have to use substitutions to deduce that for all $ n\geq 1$, that $$\int^{\pi/2}_0(\cos\theta)^{2n+1}d\theta \leq \frac{1}{\sqrt{n}}\int^{\sqrt{n}}_0e^{-y^2}dy\leq\int^{\pi/4}_0(\cos\theta)^{2n-2}d\theta$$

I have made appropriate substitutions for the first 2 integrals, but i'm struggling with the third one. I know that I have to make a transformation, for which $\theta=\pi/4 \Rightarrow u=1$. So one of the transformations appropriate for this is $\tan(\theta)$. But then when I do this, I get:

$$\int^{\pi/4}_0\left(\frac{\sin(\theta)}{\tan(\theta)}\right)^{2n-2}=$$

$u=\tan(\theta) \Rightarrow du=\frac{1}{\cos^2(\theta)}d\theta$

$$=\int^1_0\frac{\sin(\theta)^{2n-1}}{u^{2n-2}}\cos^2(\theta)d\theta=\int^1_0\frac{\sin^2(\theta)^{n-1}-\sin(\theta)^{2n}}{u^{2n-2}}d\theta$$

Well, i'm not sure what to do with this intergal. I can't see how i'm supposed to get it into the form of $\int^1_0\frac{1}{(1+x^2)^n}dx$

I also used subsition $u=\frac{\sin(\theta)}{\sqrt{2}}$ but to no avail.

Is it possible that it's a typo? It's supposed to be $\pi/2$? (the upper bound of the integral)?

If anyone could help me, that would be great!

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  • $\begingroup$ It can be expressed as a Gauss hypergeometric function $$ {}_2F_1 \left( {\tfrac{1}{2},n;\tfrac{3}{2}; - 1} \right).$$ If the upper limit of integration was $+\infty$, you would have a special case of the beta function. $\endgroup$ – Gary Apr 25 '20 at 17:33
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    $\begingroup$ The question is to prove certain inequalities$$\int^1_0(1-x^2)^ndx\leq\int^1_0e^{-nx^2}dx\leq\int^1_0\frac{1}{(1+x^2)^n}dx ,$$ which does not involve evaluation of the integrals. $\endgroup$ – GEdgar Apr 25 '20 at 20:04
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There's no typo, you just need a trigonometric identity. With $x=\tan\theta$ so that $1+x^2=\sec^2\theta$ (in fact this is the motive for our choice of substitution),$$\int_0^1\frac{dx}{(1+x^2)^n}=\int_0^{\pi/4}\frac{\sec^2\theta d\theta}{(\sec^2\theta)^n}=\int_0^{\pi/4}\cos^{2n-2}\theta d\theta.$$If you wanted to deduce a suitable substitution going the other way, Bioche's rules would guide you.

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Simply use $\,\dfrac{\sin \theta}{\tan \theta} = \cos \theta.$

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  • $\begingroup$ I did, and i'm getting a the integral i showed from which i can't show anything $\endgroup$ – user650626 Apr 25 '20 at 19:47
  • $\begingroup$ I don't understand. You start with $$\int_0^1 \frac{1}{(1+x^2)^n}\,dx.$$ Then you let $x= \tan \theta$ to get $$\int^{\pi/4}_0\left(\frac{\sin \theta}{\tan\theta}\right)^{2n-2}\,d\theta =\int^{\pi/4}_0(\cos \theta)^{2n-2}\,d\theta.$$ $\endgroup$ – zhw. Apr 25 '20 at 19:55
  • $\begingroup$ i understand, but how do i do it the other way round, starting from $\int^{\pi/4}_0(\cos \theta)^{2n-2}\,d\theta.$? $\endgroup$ – user650626 Apr 25 '20 at 20:00
  • $\begingroup$ @Marina Why so?! Your task is to start with $x $ and finish with $\theta $, not other way round. $\endgroup$ – user Apr 25 '20 at 20:06
  • $\begingroup$ @user well for the first one i did: $\int^{\pi/2}_0(\cos(\theta)^{2n+1}=\int^{\pi/2}_0(1-sin^2(\theta)^{\frac{1}{2}2n+1}=\int^{\pi}_0(1-sin^2\theta)^{\frac{2n+1}{2}}$ Substitution: $sin(\theta)=u) \Rightarrow \cos(\theta)d\theta=du$ so the integral is $\int^1_0\frac{(1-u^2)^{\frac{2n+1}{2}}}{\cos(\theta)}=\int^1_0\frac{(1-u^2)^{\frac{2n+1}{2}}}{\sqrt{1-u^2}}du=\int^1_0(1-u^2)^ndu$ So this is how i get it from the integral to the form i need, and that's why i asked for that method. At first i didnt think of doing it the other way round, which is easier(or not) in the last case $\endgroup$ – user650626 Apr 25 '20 at 20:15
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$$\int_0^1 \frac {dx}{(1+x^2)^n}\stackrel{x=\tan t}=\int_0^{\pi/4} (\cos^2t)^n\frac {dt}{\cos^2 t}=? $$

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  • $\begingroup$ but how would I get the form $\frac{1}{(1+x^2)^n}$ from $\cos(\theta)^{2n-2}$? $\endgroup$ – user650626 Apr 25 '20 at 19:58
  • $\begingroup$ It is other way around: you get $\cos (\theta)^{2n}$ from $\frac1 {(1+x^2)^n}$ $\endgroup$ – user Apr 25 '20 at 20:01

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