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$$\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin x|}{1+\pi^x}\,dx=a((n+b)!-c!),$$where $a,b,c\in \Bbb N$. Find the value of $a+b+c$.

My attempt

Let $$I=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin x|}{1+\pi^x}\,dx.\tag{1}$$

Apply $\int\limits_{a}^{b}f(x)\,dx=\int\limits_{a}^{b}f(a+b-x)\,dx$ to get

$$I=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin(-x)|}{1+\pi^{-x}}\,dx.\tag{2}$$

Add $(1)$ and $(2)$,

$$2I=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\left| \sin x\right|\,dx=2\sum_{r=1}^n\int_{0}^{r(r!)}| \sin x|\,dx$$

$$I=\sum_{r=1}^n\int_{0}^{r(r!)}\sin x \,dx$$

I could not solve further and factorials in the answer are confusing me.

Clayton
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Vinod Kumar Punia
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    I fixed up your formatting; you might take a look so you can learn how to post questions that are more visually appealing. Also, I think you have missed absolute values around the final integral. – Clayton Jul 28 '15 at 16:27
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    Are you really sure the integral is with $|\sin x|$ and not with $|\sin(\pi x)|$? – Jack D'Aurizio Jul 28 '15 at 17:02
  • @JackD'Aurizio,yes Sir,it is $\left|sinx\right|$.This is what written in the book from where i have taken this question. – Vinod Kumar Punia Jul 29 '15 at 02:11
  • Why down-vote,what i have down wrong?The question is correct to the best of my knowledge. – Vinod Kumar Punia Jul 29 '15 at 02:12
  • What happened to the denominators when you added (1) and (2)? – user170231 Aug 14 '15 at 04:46
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    It's quite common when I see $n\cdot n!$ that the solution involves noting that: $$n\cdot n!=(n+1)!-n!.$$ Not sure if this helps here. – Thomas Andrews Aug 14 '15 at 11:30
  • Numerically evaluating $I$ when $n=1$ provides $I_1 \approx 0.0846.\ $ You (or the book) are almost certainly missing a $\pi$ somewhere. – will Aug 15 '15 at 00:42
  • $2I=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin x|}{1+\pi^x}+\frac{|\sin x|}{1+\pi^{-x}}dx=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin x|}{1+\pi^x}+\frac{\pi^x|\sin x|}{1+\pi^{x}}dx=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{(1+\pi^x)|\sin x|}{1+\pi^x}=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}|\sin x|dx$ – Vinod Kumar Punia Aug 16 '15 at 15:44

1 Answers1

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The problem as it stands is not consistent. If we take the case $n=1$ we get

$$\int_{-1}^1\frac{|\sin(x)|}{1+\pi^x}\,{\rm d}x = a((b+1)!-c!)$$

The right hand side is supposed to be an integer while the left hand side evaluates to $\simeq 0.4596$.

We can make the problem consistent by making a very simple change, namely the replacement $$r(r!) \to \pi r(r!)$$ This makes it plausible that there is a typo and that this is the 'true' problem that the problem-creator meant to ask. However, there might be other ways to do it so I can't say for sure. For example changing $\sin(x)\to \pi\sin(\pi x)$ which was mentioned by Jack in the comments above also works and is in fact equivalent to the $\pi r(r!)$ replacement.

With the change $r(r!) \to \pi r (r!)$ to the problem statement it becomes easy to solve since you have already done the tricky part of this problem which is to use a clever substitution to reduce the sum down to $\sum_{r=1}^n I(r)$ where

$$I(r) = \int_0^{\pi r(r!)}|\sin(x)|\,{\rm d} x$$

This integral evaluates to $I(r) = r(r!)\int_0^{\pi}|\sin(x)|\,{\rm d} x = 2r(r!)$ and it follows that

$$\sum_{r=1}^n I(r) = \sum_{r=1}^n 2[(r+1)! - r!] = 2[(n+1)!-1] \implies a+b+c = 4$$

Winther
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