1. ## Integral

Show that

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

2. I'd try substituting $z = r \cos \theta$ 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 ...

3. 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.

4. First, use the sum of angles formula on $\cos(x+r\sin{x})$:
$\cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})$,
so our integral becomes:
$\int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx$
$=\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$
Now, one can simply note that $\frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}$, and that $\frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}$, so that our integral is
$\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$ $=\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx$
We recogize this as having the form $u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv)$, and so:
$\int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s in(r\sin{x})\right)\,dx$
$=\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi}$
$=\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r }\right)=0$

-Kevin C.

5. Originally Posted by TwistedOne151
First, use the sum of angles formula on $\cos(x+r\sin{x})$:
$\cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})$,
so our integral becomes:
$\int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx$
$=\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$
Now, one can simply note that $\frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}$, and that $\frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}$, so that our integral is
$\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$ $=\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx$
We recogize this as having the form $u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv)$, and so:
$\int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s in(r\sin{x})\right)\,dx$
$=\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi}$
$=\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r }\right)=0$

-Kevin C.
nice! i like this integral better though: $\int_0^{\pi} e^{r \cos x} \cos (r \sin x) \ dx.$

6. Originally Posted by TwistedOne151
First, use the sum of angles formula on $\cos(x+r\sin{x})$:
$\cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})$,
so our integral becomes:
$\int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx$
$=\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$
Now, one can simply note that $\frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}$, and that $\frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}$, so that our integral is
$\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx$ $=\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx$
We recogize this as having the form $u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv)$, and so:
$\int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s in(r\sin{x})\right)\,dx$
$=\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi}$
$=\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r }\right)=0$

-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: $\int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}$, can you go ahead and find $\int_{0}^{a} e^{r\cos x}\cos(2x+r\sin x)\,dx=?$

7. Originally Posted by luobo

Now you have got this: $\int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}$, can you go ahead and find $\int_{0}^{a} e^{r\cos x}\cos(2x+r\sin x)\,dx=?$
sure! just differentiate both sides of $\int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}$ with respect to $r.$

8. Originally Posted by NonCommAlg
sure! just differentiate both sides of $\int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}$ with respect to $r.$
To me, that's the beauty of this integral. You can continue the differentiation to find $\int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx, n\geq 1$ and you can also find the curve $f(x)=e^{r\cos x}\cos(nx+r\sin x)$ is a beauty as well in $[0, \pi]$ when you plot it out.

$\int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx=\frac{d^{n-1}}{dr^{n-1}}\left(\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}\righ t), n\geq 1$

I am sure NonCommAlg has a solution for $n=0$, so can you please share with us?

Additional question is: What happens if $n$ is zero or negative. Is there a closed-form solution. I guess not, if so, how about if $a =\pi$. I don't have a solution for this. If you have, can you please share.

9. Originally Posted by NonCommAlg
nice! i like this integral better though: $\int_0^{\pi} e^{r \cos x} \cos (r \sin x) \ dx.$

Which is $\pi$, independent of $r$. I wonder if this one has a closed-form formula in terms of $r$ so that by taking derivative of it you get the case for $n=1$.

10. $\int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx=\frac{d^{n-1}}{dr^{n-1}}\left(\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}\righ t), n\geq 1$ (*)

Let me give an example showing how the previous equality is used to get the summation of an important series.

[/COLOR][/COLOR]
$f(r)=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}$ (1)

According to the equality,
$f^{(n-1)}(r)=\int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx$
(2)

Therefore,
$f^{(n-1)}(0)=\int_{0}^{a} \cos(nx)\,dx=\frac{\sin(na)}{n}$
(3)

Taylor's series,
$f(r)=\sum_{n=1}^{\infty}\frac{f^{(n-1)}(0)}{(n-1)!}\,$
$r^{n-1}$ (4)

Plug (3) into (4),
$f(r)=\sum_{n=1}^{\infty}\frac{\sin(na)}{n!}\,$ $r^{n-1}$ (5)

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

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

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

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

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

By Luobo
08/14/2009

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

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

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

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

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

and you directly get 0