What is $PQ^{-1}$ if $P=\int^{\pi}_{0}\frac{\sin(994 x)}{\sin x}\sin(1332x)\,dx$ and $Q=\int^{1}_{0}\frac{x^{338}(x^{1988}-1)}{x^2-1}\,dx$?

Question

If $\displaystyle P=\int^{\pi}_{0}\frac{\sin(994 x)}{\sin x}\cdot \sin(1332x)\,dx$ and $\displaystyle Q=\int^{1}_{0}\frac{x^{338}(x^{1988}-1)}{x^2-1}\,dx$. Then $P\cdot Q^{-1}$ is

Try: put $x=e^{i\theta}$ in $\displaystyle Q=\int^{1}_{0}\frac{x^{338}(x^{1988}-1)}{x^2-1}dx$

$\displaystyle Q =\int \frac{e^{i338\theta}(e^{i1988\theta}-1)}{e^{i2x}-1}\cdot ie^{i\theta}d\theta$

$\displaystyle Q=\int\frac{(\cos (1332x)+i\sin (1332 x))\cdot \sin(994x)}{\sin x}\cdot i dx$

Could some help me How to convert limit and solve that integral . Thanks

Possible duplicate of Ratio of $2$ definite Integrals (different constants, but otherwise exactly the same problem) — Winther, Apr 21 '19 at 11:14
Thanks Winther and TheSimpleFire. But where i am wrong in my solution. — DXT, Apr 21 '19 at 15:36
You can't substitute $x = e^{i\theta}$ as that would require $\theta$ to go from $x=0$ to $x = +i\infty$ in the complex plane (and not $0\to \pi$) to match the given integration limits. — Winther, Apr 21 '19 at 16:22
@DXT: What didn't you like about the linked post? It contains everything you need to solve your problem, which is almost the one there. — Alex M., Apr 23 '19 at 18:43
$\texttt{Mathematica}$ yields $\color{red}{\large 2}$ instead of $\color{red}{\large -1}$. — Felix Marin, Apr 24 '19 at 06:13

Yuri Negometyanov · Accepted Answer · 2019-04-28T16:10:44.640

Very nice idea!

More accurate approach is $$P = \int\limits_0^\pi \dfrac{\sin994x\sin1332x}{\sin x} \,dx = \dfrac12\int\limits_0^\pi\dfrac{\cos338x - \cos2326x}{\sin x} \,dx$$ $$ = \dfrac12\Re\int\limits_0^\pi\dfrac{e^{338ix} - e^{2326ix}}{\sin x} \,dx = \dfrac12\Re\int\limits_0^\pi\dfrac{e^{338ix} - e^{2326ix}}{e^{ix}-e^{-ix}} \,d(ix) = \Re\oint\limits_{|z|=1\\ \Im z>0}\dfrac{z^{338}-z^{2326}}{z^2-1}dz$$ $$ = \Re\lim\limits_{\delta\to 0}\left(\oint\limits_{z=-1+\delta e^{i\varphi}\\ \varphi = \frac\pi2\to 0}\dfrac{z^{2326}-z^{338}}{z^2-1}dz + \int\limits_{-1}^1\dfrac{z^{2326}-z^{338}}{z^2-1}dz + \oint\limits_{z=1+\delta e^{i\varphi}\\ \varphi = \pi\to\frac\pi2}\dfrac{z^{2326}-z^{338}}{z^2-1}dz\right)$$ $$ = \Re\lim\limits_{\delta\to 0}\left(\int\limits_{\large \frac\pi2}^0\dfrac{(-1+\delta e^{i\varphi})^{2326}-(-1+\delta e^{i\varphi})^{338}}{(-1+\delta e^{i\varphi})^2-1}d\varphi +\int\limits_\pi^{\large \frac\pi2}\dfrac{(1+\delta e^{i\varphi})^{2326}-(1+\delta e^{i\varphi})^{338}}{(1+\delta e^{i\varphi})^2-1}d\varphi\right) + \int\limits_{-1}^1\dfrac{x^{2326}-x^{338}}{x^2-1}dx$$ $$ = {\Re\lim\limits_{\delta\to 0}}\left(\int\limits_{\large \frac\pi2}^0\dfrac{2326e^{i\varphi}(-1+\delta e^{i\varphi})^{2325}-338e^{i\varphi}(-1+\delta e^{i\varphi})^{337}}{2e^{i\varphi}(-1+\delta e^{i\varphi})}d\varphi +\int\limits_\pi^{\large \frac\pi2}\dfrac{2326e^{i\varphi}(1+\delta e^{i\varphi})^{2325}-338e^{i\varphi}(1+\delta e^{i\varphi})^{337}}{2e^{i\varphi}(1+\delta e^{i\varphi})}d\varphi\right) + 2\int\limits_0^1\dfrac{(x^{1938}-1)x^{338}}{x^2-1}dx$$ $$ = \dfrac\pi4(-2326+338+2326-338) + 2Q\color{brown}{\mathbf{ = 2Q.}}$$

$\texttt{Mathematica}$ yields $\color{red}{\large 2}$ instead of $\color{red}{\large -1}$. — Felix Marin, Apr 24 '19 at 06:15

What is $PQ^{-1}$ if $P=\int^{\pi}_{0}\frac{\sin(994 x)}{\sin x}\sin(1332x)\,dx$ and $Q=\int^{1}_{0}\frac{x^{338}(x^{1988}-1)}{x^2-1}\,dx$?

1 Answers1