1. Originally Posted by Random Variable
It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

$\int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)$
Easy one. Again, the magic differentiation, with respect to a or b !!! I would let others to finish.

2. Originally Posted by Random Variable
It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

$\int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)$
we may use the result: $f$is continuous and $f(+\infty)$exists,then
$\int_0^{+\infty} \frac{f(ax)-f(bx)}{x} \ dx =(f(0)-f(+\infty))\ln \frac{b}{a}$

3. Originally Posted by Krizalid
Try this one:

Evaluate $\lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{ \sin^nx+\cos^nx}\,dx.$
$\frac{\pi}{6}$ is it right ?

We can obtain $\int_a^{\frac{\pi}{2} - a } \frac{ dx}{ 1 + \cot^n(x) }$ for all value of n

and at the same time when $0 \leq x \leq \frac{\pi}{4}$ , $\frac{1}{1 + \cot^n(x)}$ tends to 0 when n tends to a large value

But i am not sure , if i sketch the graph , It is just like a step function ..

4. No, but you're close.

5. Originally Posted by Krizalid
Try this one:

Evaluate $\lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{ \sin^nx+\cos^nx}\,dx.$
let $u = \frac{\pi}{ 2} - x$

then $du = -dx$

$\cos(\pi/2 - u) = \sin u$

$\sin(\pi/2 -u) = \cos u$

$\int^{\pi/3}_{0}\frac{\sin^{n}(x)}{\sin^{n}x + \cos^{n}x} dx = \int^{\pi/2}_{\pi/6}\frac{\cos^{n}(u)}{\sin^{n}u + \cos^{n}u} du$

at which point I realize that this method doesn't work because of the upper limit not being $\pi/2$

6. Originally Posted by Krizalid
No, but you're close.
Oh , the integral $\lim_{n\to\infty} \int_0^{\frac{\pi}{3}} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx$ and

$\int_0^{\frac{\pi}{3}} \lim_{n\to\infty} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx$

are different ?

Or oh i have made a mistake , it should be $\frac{\pi}{12}$ ?

7. Originally Posted by ynj
we may use the result: $f$is continuous and $f(+\infty)$exists,then
$\int_0^{+\infty} \frac{f(ax)-f(bx)}{x} \ dx =(f(0)-f(+\infty))\ln \frac{b}{a}$
Yep, it's a Frullani integral.

8. Originally Posted by Random Variable
Yep, it's a Frullani integral.
Proof of Frullani Integral (assuming f(x) is differentiable, which is stronger than continuous):

$
I(a)=\int_0^{\infty} \frac{f(ax)-f(bx)}{x} \ dx
$

$
\frac{dI}{da}=\int_0^{\infty} f'(ax) dx = [f(\infty)-f(0)]\frac{1}{a}
$

$
$

Obviously when $a=b,\; I=0$, i.e. $[f(\infty)-f(0)]\ln b+C=0\Rightarrow C=[f(0)-f(\infty)]\ln b\quad (2)$

Combine (1) and (2), we get the previous result.

9. Originally Posted by simplependulum
Oh , the integral $\lim_{n\to\infty} \int_0^{\frac{\pi}{3}} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx$ and

$\int_0^{\frac{\pi}{3}} \lim_{n\to\infty} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx$

are different ?

Or oh i have made a mistake , it should be $\frac{\pi}{12}$ ?
Such kind of integral, in general, after certain variable substitution (for example, $t=\tan\frac{x}{2}$), can be transformed into rational functions.

10. Originally Posted by Krizalid
Try this one:

Evaluate $\lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{ \sin^nx+\cos^nx}\,dx.$
as simplependulum mentioned (without a proof again! haha) the answer is $\frac{\pi}{12}.$ more generally if $\frac{\pi}{4}< \alpha < \frac{\pi}{2},$ and $I_n(\alpha)=\int_0^{\alpha}\frac{\sin^nx}{\sin^nx+ \cos^nx}\,dx,$ then $\lim_{n\to\infty}I_n(\alpha)=\alpha - \frac{\pi}{4}.$

the reason is that $I_n(\alpha)=\int_0^{\frac{\pi}{2}}\frac{\sin^nx}{\ sin^nx+\cos^nx}\,dx - \int_{\alpha}^{\frac{\pi}{2}}\frac{\sin^nx}{\sin^n x+\cos^nx}\,dx=\frac{\pi}{4} - \int_{\alpha}^{\frac{\pi}{2}}\frac{1}{1+\cot^nx}\, dx.$

now, since $0 \leq \cot x \leq \cot \alpha < 1$ for $\alpha \leq x \leq \frac{\pi}{2},$ we have $\lim_{n\to\infty}\int_{\alpha}^{\frac{\pi}{2}}\fra c{1}{1+\cot^nx}\,dx=\int_{\alpha}^{\frac{\pi}{2}} dx = \frac{\pi}{2} - \alpha$ and therefore $\lim_{n\to\infty} I_n(\alpha) = \alpha - \frac{\pi}{4}.$

-------------------------------------------------------------

here's a problem: Solve for real values of $x,y: \ \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt = \frac{\pi x^2}{12}.$

11. here's a problem: Solve for real values of $x,y: \ \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt = \frac{\pi x^2}{12}.$
Message deleted

12. Originally Posted by NonCommAlg
the question is to find all $(x,y) \in \mathbb{R}^2$ for which the given equality holds. well, (0,0) is just one solution.
I had thought the problem is a proof problem. It actually is equation solving. So I removed my post and you replied really fast. Are you here all day and every minute?

13. It is time for you to apply " magic differentiation "

14. and the last problem: Suppose $a,b,c > 0.$ Evaluate $\lim_{n\to\infty} \frac{1}{n^3} \int_1^e \ln(1 + ax^n) \ln(1 + bx^n) \ln(1 + cx^n) \ dx.$

15. Sorry , i forgot it , now let me post the solution to
$\int_0^{\pi} (\sin(x))^a \sin(ax)~dx$

consider

$(1/2i)^a \int_0^{\pi} e^{iax} ( e^{ix} - e^{-ix} )^a ~dx$

$= (1/2i)^a \int_0^{\pi} ( e^{2ix} - 1 )^a ~dx$

Sub $2x = t$

$= (1/2i)^a (1/2) \int_0^{2\pi} ( e^{it} - 1)^a ~ dt$

adn sub $z = e^{it}$

the integral becomes

$(1/2i)^{a} (1/2i) (-1)^a \int_{|z|=1} \frac{ (1-z)^a}{z} ~dz$

$= (1/2)^a e^{ai(\pi - \frac{\pi}{2})} (1/2i) ( 2\pi i )$

$= \frac{\pi}{2^a} e^{i \frac{\pi}{2} }$

Oh there is something wrong : the last line , it should be $= \frac{\pi}{2^a} e^{i \frac{a\pi}{2} }$

so $\int_0^{\pi} e^{iax} ( \sin(x))^a ~dx = \frac{\pi}{2^a} e^{i \frac{\pi}{2} }$

compare the Im part $\int_0^{\pi} \sin(ax) ( \sin(x))^a ~dx = \frac{\pi}{2^a} \sin(a\frac{\pi}{2})$
( note : we can see $\sin(x) \geq 0 , 0 \leq x \leq \pi$ then $(\sin(x))^a \in \mathbb{R^+}$ so that we can compare the real and imaginary part

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