# Integral

• Aug 14th 2009, 11:14 AM
luobo
Integral
Show that

$\displaystyle \int_{0}^{\pi} e^{r\cos x}\cos(x+r\sin x)\,dx=0$
• Aug 14th 2009, 11:24 AM
Matt Westwood
I'd try substituting $\displaystyle z = r \cos \theta$ and see where that got me. What have you tried so far?

Sorry, ignore me - I didn't notice this was a challenge forum! Okay, I'll have a go ...
• Aug 14th 2009, 12:22 PM
Matt Westwood
Hang on, this is Bessel functions, isn't it? Way over my head, sorry about that. I'm going to have to give this one a miss till I've learned up on it.
• Aug 14th 2009, 12:30 PM
TwistedOne151
First, use the sum of angles formula on $\displaystyle \cos(x+r\sin{x})$:
$\displaystyle \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})$,
so our integral becomes:
$\displaystyle \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx$
$\displaystyle =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$
Now, one can simply note that $\displaystyle \frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}$, and that $\displaystyle \frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}$, so that our integral is
$\displaystyle \int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$$\displaystyle =\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx We recogize this as having the form \displaystyle u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv), and so: \displaystyle \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s in(r\sin{x})\right)\,dx \displaystyle =\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi} \displaystyle =\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r }\right)=0 -Kevin C. • Aug 14th 2009, 12:47 PM NonCommAlg Quote: Originally Posted by TwistedOne151 First, use the sum of angles formula on \displaystyle \cos(x+r\sin{x}): \displaystyle \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x}), so our integral becomes: \displaystyle \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx \displaystyle =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx Now, one can simply note that \displaystyle \frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}, and that \displaystyle \frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}, so that our integral is \displaystyle \int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$$\displaystyle =\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx$
We recogize this as having the form $\displaystyle u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv)$, and so:
$\displaystyle \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s in(r\sin{x})\right)\,dx$
$\displaystyle =\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi}$
$\displaystyle =\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r }\right)=0$

-Kevin C.

nice! i like this integral better though: $\displaystyle \int_0^{\pi} e^{r \cos x} \cos (r \sin x) \ dx.$
• Aug 14th 2009, 01:03 PM
luobo
Quote:

Originally Posted by TwistedOne151
First, use the sum of angles formula on $\displaystyle \cos(x+r\sin{x})$:
$\displaystyle \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})$,
so our integral becomes:
$\displaystyle \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx$
$\displaystyle =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$
Now, one can simply note that $\displaystyle \frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}$, and that $\displaystyle \frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}$, so that our integral is
$\displaystyle \int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$$\displaystyle =\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx We recogize this as having the form \displaystyle u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv), and so: \displaystyle \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s in(r\sin{x})\right)\,dx \displaystyle =\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi} \displaystyle =\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r }\right)=0 -Kevin C. That's right. I just happened to find out this interesting integral when I was doing the line integral posted here today by transgalactic, using both line integral and double integral. http://www.mathhelpforum.com/math-he...-question.html Now you have got this: \displaystyle \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}, can you go ahead and find \displaystyle \int_{0}^{a} e^{r\cos x}\cos(2x+r\sin x)\,dx=? • Aug 14th 2009, 01:08 PM NonCommAlg Quote: Originally Posted by luobo Now you have got this: \displaystyle \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}, can you go ahead and find \displaystyle \int_{0}^{a} e^{r\cos x}\cos(2x+r\sin x)\,dx=? sure! just differentiate both sides of \displaystyle \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r} with respect to \displaystyle r. (Evilgrin) • Aug 14th 2009, 01:19 PM luobo Quote: Originally Posted by NonCommAlg sure! just differentiate both sides of \displaystyle \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r} with respect to \displaystyle r. (Evilgrin) To me, that's the beauty of this integral. You can continue the differentiation to find \displaystyle \int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx, n\geq 1 and you can also find the curve \displaystyle f(x)=e^{r\cos x}\cos(nx+r\sin x) is a beauty as well in \displaystyle [0, \pi] when you plot it out. \displaystyle \int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx=\frac{d^{n-1}}{dr^{n-1}}\left(\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}\righ t), n\geq 1 I am sure NonCommAlg has a solution for \displaystyle n=0, so can you please share with us? Additional question is: What happens if \displaystyle n is zero or negative. Is there a closed-form solution. I guess not, if so, how about if \displaystyle a =\pi. I don't have a solution for this. If you have, can you please share. • Aug 14th 2009, 03:37 PM luobo Quote: Originally Posted by NonCommAlg nice! i like this integral better though: \displaystyle \int_0^{\pi} e^{r \cos x} \cos (r \sin x) \ dx. Which is \displaystyle \pi , independent of \displaystyle r . I wonder if this one has a closed-form formula in terms of \displaystyle r so that by taking derivative of it you get the case for \displaystyle n=1. • Aug 14th 2009, 06:50 PM luobo \displaystyle \int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx=\frac{d^{n-1}}{dr^{n-1}}\left(\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}\righ t), n\geq 1 (*) Let me give an example showing how the previous equality is used to get the summation of an important series. [/COLOR][/COLOR] \displaystyle f(r)=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r} (1) According to the equality, \displaystyle f^{(n-1)}(r)=\int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx (2) Therefore, \displaystyle f^{(n-1)}(0)=\int_{0}^{a} \cos(nx)\,dx=\frac{\sin(na)}{n} (3) Taylor's series, \displaystyle f(r)=\sum_{n=1}^{\infty}\frac{f^{(n-1)}(0)}{(n-1)!}\, \displaystyle r^{n-1} (4) Plug (3) into (4), \displaystyle f(r)=\sum_{n=1}^{\infty}\frac{\sin(na)}{n!}\,$$\displaystyle r^{n-1}$ (5)

Multiply $\displaystyle r$ on both sides,
$\displaystyle rf(r)=\sum_{n=1}^{\infty}\frac{\sin(na)}{n!}\,$$\displaystyle r^{n}$ (6)

Finally, the series:
$\displaystyle \sum_{n=1}^{\infty}\frac{\sin(na)}{n!}\,r^{n}=e^{r \cos a}\sin(r\sin a)$ (7)

Some Special cases:
(a) $\displaystyle r=1$
$\displaystyle \sum_{n=1}^{\infty}\frac{\sin(na)}{n!}=e^{\cos a}\sin(\sin a)$ (8)

(b) $\displaystyle r=1, a=1$
$\displaystyle \sum_{n=1}^{\infty}\frac{\sin n}{n!}=e^{\cos 1}\sin(\sin1)$ (9)

(c) $\displaystyle a=\frac{\pi}{2}$
$\displaystyle \sum_{n=1}^{\infty}\frac{\sin (n\pi/2)}{n!}r^n=\sin r$ (10) (which is within everybody's expectation)

By Luobo
08/14/2009
• Aug 14th 2009, 06:53 PM
pankaj
Quote:

Originally Posted by TwistedOne151
First, use the sum of angles formula on $\displaystyle \cos(x+r\sin{x})$:
$\displaystyle \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})$,
so our integral becomes:
$\displaystyle \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx$
$\displaystyle =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$

-Kevin C.

Consider $\displaystyle \int_0^{\pi}e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$.

Integrating by parts
$\displaystyle \int_0^{\pi}e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$

$\displaystyle =\frac{1}{r}\left[-e^{r\cos x}\sin (r\sin x)\right]_{x=0}^{\pi}-\frac{1}{r}\left[\int_{0}^{\pi}(-e^{r\cos x}).\cos(r\sin x).r\cos x.dx\right]$

$\displaystyle =\int_{0}^{\pi}e^{r\cos x}.\cos(r\sin x).\cos x.dx$

and you directly get 0(Wink)