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!


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $$\frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}x\,dx}{\int_0^{\pi/2}\sin^{\sqrt{2}-1}x\,dx}=\frac{B(1+\frac{\sqrt{2}}{2},\frac12)}{B(\frac{\sqrt{2}}{2},\frac12)}=\frac{\Gamma(1+\frac{\sqrt{2}}{2})\Gamma(\frac{1+\sqrt{2}}{2})}{\Gamma(\frac{\sqrt{2}}{2})\Gamma(\frac{3+\sqrt{2}}{2})}=\frac{\sqrt{2}}{1+\sqrt{2}}=2-\sqrt{2}.$$

  • $\begingroup$ 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

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*}

  • 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

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