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Let $\alpha,\beta$ be the distinct positive roots of the equation $\tan x=2x$,then find $\int_{0}^{1}\sin \alpha x \sin \beta x$dx,independent of $\alpha$ and $\beta$.

My Attempt

$\tan \alpha=2\alpha$

$\tan \beta=2\beta$

Adding the two equations,we get $2\alpha+2\beta=\frac{\sin (\alpha+\beta)}{\cos \alpha\cos \beta}$.....(1)

$\int_{0}^{1}\sin \alpha x \sin \beta x$dx$=\int_{0}^{1}\sin (\alpha-\alpha x) \sin (\beta-\beta x)$dx

Then i got stuck,some hints are required.

user1442
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3 Answers3

4

You can do simpler writing $$\sin(\alpha x)\sin(\beta x)=\frac 12 \Big(\cos\big((\alpha-\beta)x\big)-\cos\big((\alpha+\beta)x\big)\Big)$$ So $$A=\int_0^1 \sin(\alpha x)\sin(\beta x)\,dx=\frac 12 \Big(\frac{\sin(\alpha-\beta)}{\alpha-\beta}- \frac{\sin(\alpha+\beta)}{\alpha+\beta}\Big)$$ Now, reduce to same denominator and expand the sines; you should arrive to $$A=\frac{\beta \sin (\alpha ) \cos (\beta )-\alpha \cos (\alpha ) \sin (\beta )}{\alpha ^2-\beta ^2}$$ Factor now $\cos (\alpha ) \cos (\beta )$ to get $$A=\frac{\cos (\alpha ) \cos (\beta )}{\alpha ^2-\beta ^2} \big(\beta \tan(\alpha)-\alpha \tan(\beta)\big)$$ But you know that $\tan(\alpha)=2\alpha$ and $\tan(\beta)=2\beta$.

I am sure that you can take from here.

Claude Leibovici
  • 260,315
4

You already have nice solutions, but one might wonder where such a problem comes from. This answer provides a setting of differential equations. Once it is put in this setting, the result of the integral follows from a general theorem.

When solving the Sturm-Liouville eigenvalue problem $$ -u''(t)=\lambda u(t),\quad u(0)=0,\ 2u'(1)-u(1)=0, $$ one first finds that the general solution to the differential equation is $$ u(t)=A\cos\sqrt{\lambda} t+B\sin\sqrt{\lambda}t. $$ Imposing the first boundary condition, the cosine term disappears, and one get $$ u(t)=B\sin\sqrt{\lambda}t. $$ Now, the second boundary condition implies that $$ 0=2u'(1)-u(1)=2B\sqrt{\lambda}\cos\sqrt{\lambda}-B\sin\sqrt{\lambda}. $$ Looking for non-trivial solutions, $B\neq 0$, and we find that $\lambda$ must satisfy $$ \tan\sqrt{\lambda}=2\sqrt{\lambda} $$ This equation has infinitely many positive solutions, denote them by $\lambda_k$, $k=1$, $2$, $\ldots$. These are called eigenvalues of the Sturm-Liouville problem above. The corresponding functions $u_k(t)=\sin\sqrt{\lambda_k}t$ are called eigenfunctions.

There is a very nice theorem that fits in this setting:

Theorem Eigenfunctions corresponding to different eigenvalues are orthogonal.

In our case, orthogonal means that $$ \int_0^1 u_j(t) u_k(t)\,dt=0\quad\text{if }j\neq k. $$

With $\alpha=\sqrt{\lambda_j}$ and $\beta=\sqrt{\lambda_k}$, it follows that your integral is zero.

mickep
  • 19,962
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Given $\displaystyle \tan x= 2x.$ Given $\alpha$ and $\beta$ are two roots of $\tan x= 2x.$

So we get $2\alpha = \tan \alpha$ and $2\beta = \tan \beta$

Now Given $$\displaystyle I = \int_{0}^{1}\sin \alpha x\cdot \sin \beta xdx =\frac{1}{2}\int_{0}^{1}2\sin \alpha x\cdot \sin \beta xdx $$

$\displaystyle I = \frac{1}{2}\int_{0}^{1}\left[\cos (\alpha -\beta)x-\cos (\alpha+\beta)x\right]dx.$

$$\displaystyle I = \frac{1}{2(\alpha-\beta)}[\sin(\alpha-\beta )x]_{0}^{1}-\frac{1}{2(\alpha+\beta)}[\sin(\alpha-\beta )x]_{0}^{1}=\frac{\sin (\alpha-\beta)}{2(\alpha-\beta)}-\frac{\sin (\alpha+\beta)}{2(\alpha+\beta)}$$

Now From above Values of $2\alpha$ and $2\beta $ from above.

so we get $$2(\alpha-\beta) = \tan \alpha -\tan \beta$$ and $$2(\alpha+\beta) = \tan \alpha +\tan \beta$$

Put into Integral $I$ . So we get $$\displaystyle I = \frac{\sin (\alpha-\beta)}{\tan \alpha -\tan \beta}-\frac{\sin (\alpha+\beta)}{\tan \alpha +\tan \beta} = 0 $$

juantheron
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