15
$\begingroup$

I have this difficult integral to solve.

$$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $$

Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^{\sqrt 2 - 1}$ as $(\sin x)^{\sqrt 2}.(\sin x)$ and $(\sin x)^{\sqrt 2 - 2}.(\sin x)$ respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please!

Edit:

This is what I did (showing just for the numerator)

$$ \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx $$ $$ = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx $$ $$ = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 }(\cos^2 x) dx $$

(taking $ v = \sin x $ and $ u = (\sin x)^{\sqrt 2} $ in the $ \int uv $ formula)

$$ = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } \right) dx $$

Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer.

$\endgroup$
  • $\begingroup$ I don't know anything about complex integration. Is this problem solvable without using complex numbers or gamma functions? $\endgroup$ – Parth Thakkar Jul 17 '13 at 14:26
23
$\begingroup$

Let the top integral be $I$, and the bottom one $J$. To make typing simpler, let $a=\sqrt{2}$.

Integrate the top one by parts, letting $du=\sin x$ and $v=\sin^{a} x$. This is the standard way to get a reduction formula for $\int \sin^n x\,dx$.

So $dv=a\cos x\sin^{a-1}x\,dx$ and we can take $u=-\cos x$. Then $$I=\left. -\cos x \sin^{a}x\right|_0^{\pi/2}+\int_0^{\pi/2}a\cos^2 x\sin^{a-1} x\,dx.$$ The first part dies at both ends. Rewrite $\cos^2 x$ as $1-\sin^2 x$. Then $$I=aJ -aI.$$ Now we get $$I=\frac{a}{a+1}J$$ and it's over.

$\endgroup$
  • $\begingroup$ As I said, that's what I did. But it didn't help. $\endgroup$ – Parth Thakkar Jul 17 '13 at 14:27
  • $\begingroup$ OK, I will write out the details. $\endgroup$ – André Nicolas Jul 17 '13 at 14:29
  • $\begingroup$ I don't understand just one last part. How did you get $I=\sqrt{2}J -\sqrt{2}I$ ? Because I am getting $I=J-I$. Where did the $\sqrt 2$ come from? $\endgroup$ – Parth Thakkar Jul 17 '13 at 14:53
  • $\begingroup$ I still don't get the $aJ - aI$ part. My last step was $$ I = \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } dx = J - I$$ and that's what you did too. But then how did $a$ (or $\sqrt 2$ suddenly appear? I think I am making a really dumb mistake but I don't get what it is. $\endgroup$ – Parth Thakkar Jul 17 '13 at 14:57
  • $\begingroup$ We had let $v=\sin^{\sqrt{2}}x$. So when we differentiate, we get $\sqrt{2}\cos x\sin^{\sqrt{2}-1}x$. $\endgroup$ – André Nicolas Jul 17 '13 at 14:58
18
$\begingroup$

Hint: $$\int_0^{\pi/2}\sin^{2p-1}x\,\cos^{2q-1} x\,dx=\frac{1}{2}B(p,q)=\frac{\Gamma(p)\Gamma(q)}{2\Gamma(p+q)}.$$ Using this formula, we obtain $$\frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}x\,dx}{\int_0^{\pi/2}\sin^{\sqrt{2}-1}x\,dx}=\frac{B(1+\frac{\sqrt{2}}{2},\frac12)}{B(\frac{\sqrt{2}}{2},\frac12)}=\frac{\Gamma(1+\frac{\sqrt{2}}{2})\Gamma(\frac{1+\sqrt{2}}{2})}{\Gamma(\frac{\sqrt{2}}{2})\Gamma(\frac{3+\sqrt{2}}{2})}=\frac{\sqrt{2}}{1+\sqrt{2}}=2-\sqrt{2}.$$

$\endgroup$
  • $\begingroup$ Downvote??? Explain please $\endgroup$ – Start wearing purple Jul 17 '13 at 14:22
  • $\begingroup$ We haven't been taught anything about the $\Gamma$ function. Is there any alternative way? $\endgroup$ – Parth Thakkar Jul 17 '13 at 14:22
  • $\begingroup$ It wasn't me. I didn't downvote. $\endgroup$ – Parth Thakkar Jul 17 '13 at 14:23
  • $\begingroup$ (+1) I'm surprised at downvote, if it does not deserve an upvote then at least it does not also deserve a dowvote since it's a correct answer. $\endgroup$ – user63181 Jul 17 '13 at 14:30
  • $\begingroup$ @SamiBenRomdhane Thanks. Actually, I am sure that the integrals in the numerator and denominator cannot be expressed without gamma functions. Now that we have a ratio can make possible some tricks where you don't have to really calculate the integrals, as suggested by Andre Nicolas. $\endgroup$ – Start wearing purple Jul 17 '13 at 14:35
6
$\begingroup$

Just like what @O.L pointed out, you can use Beta function to express the solution. In fact, you can change variables to get \begin{eqnarray*} &&\frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}xdx}{\int_0^{\pi/2}\sin^{\sqrt{2}-1}xdx}=\frac{\int_0^{1}u^{\sqrt{2}+1}(1-u^2)^{-1/2}du}{\int_0^{1}u^{\sqrt{2}-1}(1-u^2)^{-1/2}du}\\ &=&\frac{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}+1)}(1-t)^{-1/2}\frac{1}{2t^{1/2}}dt}{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}-1)}(1-t)^{-1/2}\frac{1}{2t^{1/2}}dt}=\frac{\int_0^{1}t^{\frac{1}{2}\sqrt{2}}(1-t)^{-1/2}dt}{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}-2)}(1-t)^{-1/2}dt}\\ &=&\frac{B(\frac{\sqrt{2}+2}{2},\frac{1}{2})}{B(\frac{\sqrt{2}}{2},\frac{1}{2})}=\frac{\Gamma(\frac{\sqrt{2}+2}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{\sqrt{2}+3}{2})}\frac{\Gamma(\frac{\sqrt{2}+1}{2})}{\Gamma(\frac{\sqrt{2}}{2})\Gamma(\frac{1}{2})}=\frac{\Gamma(\frac{\sqrt{2}+2}{2})\Gamma(\frac{\sqrt{2}+1}{2})}{\Gamma(\frac{\sqrt{2}+3}{2})\Gamma(\frac{\sqrt{2}}{2})}\\ &=&\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}+1}{2}}=2-\sqrt{2}. \end{eqnarray*}

$\endgroup$
  • 3
    $\begingroup$ In view of another downvote, somebody definitely dislikes gamma functions. $\endgroup$ – Start wearing purple Jul 17 '13 at 14:57
  • $\begingroup$ Why repeat OLs answer? $\endgroup$ – Pedro Tamaroff Jul 17 '13 at 15:01
  • $\begingroup$ @PeterTamaroff At the point xpaul posted his answer, my answer contained only a hint (use beta function). I have expanded it after getting a downvote in the next second after posting it. $\endgroup$ – Start wearing purple Jul 17 '13 at 15:09
  • $\begingroup$ @O.L. Ah, I see. $\endgroup$ – Pedro Tamaroff Jul 17 '13 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.