1. Originally Posted by NonCommAlg
and the last problem: Suppose $\displaystyle a,b,c > 0.$ Evaluate $\displaystyle \lim_{n\to\infty} \frac{1}{n^3} \int_1^e \ln(1 + ax^n) \ln(1 + bx^n) \ln(1 + cx^n) \ dx.$
HaHa , we can apply magic differentiation in this problem Again !

2. Originally Posted by simplependulum
$\displaystyle \int_0^{\pi} (\sin(x))^a \sin(ax)~dx = \frac{\pi}{2^a} e^{i \frac{\pi}{2} }$
Note that $\displaystyle e^{i \frac{\pi}{2} }=i$, so you're saying
$\displaystyle \int_0^{\pi} (\sin(x))^a \sin(ax)~dx = \frac{\pi i}{2^a}$ which is a surprising result for a real integral!

3. Originally Posted by luobo
Let me have a try. . I may be wrong. If so, please criticize.

$\displaystyle \sin^2(xt) = \frac{1-\cos(2xt)}{2}$

$\displaystyle \sin^3(yt) = \frac{3\sin(yt)-\sin(3yt)}{4}$

$\displaystyle I(x)= \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt=\frac{1}{8} \int_0^\infty \frac{[1-\cos(2xt)][3\sin(yt)-\sin(3yt)]}{t^3}\;dt$

$\displaystyle \frac{dI}{dx}=\frac{1}{4} \int_0^\infty \frac{\sin(2xt)[3\sin(yt)-\sin(3yt)]}{t^2}\;dt$

$\displaystyle \frac{d^2I}{dx^2}=\frac{1}{2} \int_0^\infty \frac{\cos(2xt)[3\sin(yt)-\sin(3yt)]}{t}\;dt\quad$ (Note: second-order!!!)

$\displaystyle \frac{d^2I}{dx^2}= \frac{3}{4} \int_0^\infty \frac{\sin[(y+2x)t]+\sin[(y-2x)t]}{t}\;dt- $$\displaystyle \frac{1}{4} \int_0^\infty \frac{\sin[(3y+2x)t]+\sin[(3y-2x)t]}{t}\;dt\quad \displaystyle \frac{d^2I}{dx^2}= \frac{3}{4} \int_0^\infty \frac{\sin[(y+2x)t]}{t}\;dt+ \frac{3}{4} \int_0^\infty \frac{\sin[(y-2x)t]}{t}\;dt-$$\displaystyle \frac{1}{4} \int_0^\infty \frac{\sin[(3y+2x)t]}{t}\;dt- \frac{1}{4} \int_0^\infty \frac{\sin[(3y-2x)t]}{t}\;dt$

$\displaystyle \frac{d^2I}{dx^2}= \frac{3\pi}{8} [sign(y+2x)+sign(y-2x)] - \frac{\pi}{8} [sign(3y+2x)+sign(3y-2x)]$

Since the original integration is even of $\displaystyle x$, so can assume $\displaystyle x\geq0$ (i.e, if x is a solution, so is -x). If so, there are totally five cases in the right half plane:
(1) $\displaystyle -\infty < y \leq -2x, \quad \frac{d^2I}{dx^2}=-\frac{\pi}{2}$
(2) $\displaystyle -2x < y \leq -\frac{2}{3}x, \quad \frac{d^2I}{dx^2}=\frac{\pi}{4}$
(3) $\displaystyle -\frac{2}{3}x < y \leq \frac{2}{3}x, \quad \frac{d^2I}{dx^2}=0$
(4) $\displaystyle \frac{2}{3}x <y \leq 2x, \quad \frac{d^2I}{dx^2}=-\frac{\pi}{4}$
(5) $\displaystyle 2x < y< +\infty, \quad \frac{d^2I}{dx^2}=\frac{\pi}{2}$

The last equation tells me under a certain zone, the second-order derivative $\displaystyle \frac{d^2I}{dx^2}$ is a constant, although not continuous.

For simplicity, assume $\displaystyle \frac{d^2I}{dx^2}=a$, then $\displaystyle \frac{dI}{dx}=ax+b$ and $\displaystyle I=\frac{1}{2}ax^2+bx+c$

Obviously
$\displaystyle b=0$ since $\displaystyle \frac{dI}{dx}=0\;@\;x=0$
$\displaystyle c=0$ since $\displaystyle I=0\;@\;x=0$.

Therefore,
$\displaystyle I=\frac{1}{2}ax^2=\frac{\pi x^2}{12}$

So $\displaystyle x=0$ seems to be the only solution.
Sorry, this will not go through.

4. luobo : i am looking forward to seeing your solution , actually , I have been shocked to see how you (or someone ) applied magic differentiation in the Integral Bee !! It is really a powerful tool

i think the solution is two straight lines passing throguh the origin (at,bt) and (-at,bt)

5. Originally Posted by NonCommAlg
and the last problem: Suppose $\displaystyle a,b,c > 0.$ Evaluate $\displaystyle \lim_{n\to\infty} \frac{1}{n^3} \int_1^e \ln(1 + ax^n) \ln(1 + bx^n) \ln(1 + cx^n) \ dx.$
Simply note that: $\displaystyle \log \left( {1 + a \cdot x^n } \right) - \log \left( {a \cdot x^n } \right) = \log \left( {1 + \tfrac{1} {{a \cdot x^n }}} \right) = O\left( {\tfrac{1} {{a \cdot x^n }}} \right)$

Hence: $\displaystyle \log \left( {1 + a \cdot x^n } \right) = n \cdot \log \left( x \right) + O\left( 1 \right)$ as n tends to infinity. - x in [1, e] -

And so: $\displaystyle \log \left( {1 + a \cdot x^n } \right)\log \left( {1 + b \cdot x^n } \right)\log \left( {1 + c \cdot x^n } \right) = n^3 \cdot \log ^3 \left( x \right) + O\left( {n^2 } \right)$ - just multiply out-

Thus: $\displaystyle \int_1^e {\log \left( {1 + a \cdot x^n } \right)\log \left( {1 + b \cdot x^n } \right)\log \left( {1 + c \cdot x^n } \right)dx} = n^3 \cdot \int_1^e {\log ^3 \left( x \right)dx} + O\left( {n^2 } \right)$

And so the answer turns out to be: $\displaystyle \int_1^e {\log ^3 \left( x \right)dx} = 6 - 2e$ -by parts-

6. Here's another solution to my proposed problem using a powered tool:

$\displaystyle \int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{\frac{dx}{1+\cot ^{n}x}}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\frac{dx}{1+\cot ^{n}x}}.$

Now since $\displaystyle \frac1{1+\cot^nx}\le1$ and $\displaystyle g(x)=1$ is an integrable function on $\displaystyle x\in \left[ 0,\frac{\pi }{3} \right],$ besides on $\displaystyle \left( 0,\frac{\pi }{4} \right]$ we get $\displaystyle \frac1{1+\cot^nx}\to0$ and on $\displaystyle \left[ \frac{\pi }{4},\frac{\pi }{3} \right]$ we have $\displaystyle \frac1{1+\cot^nx}\to1,$ hence the Dominated Convergence Theorem applies and $\displaystyle \underset{n\to \infty}{\mathop{\lim }}\,\int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{0\,dx}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{1\,dx}=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}.$

7. Originally Posted by Krizalid
Here's another solution to my proposed problem using a powered tool:

$\displaystyle \int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{\frac{dx}{1+\cot ^{n}x}}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\frac{dx}{1+\cot ^{n}x}}.$

Now since $\displaystyle \frac1{1+\cot^nx}\le1$ and $\displaystyle g(x)=1$ is an integrable function on $\displaystyle x\in \left[ 0,\frac{\pi }{3} \right],$ besides on $\displaystyle \left( 0,\frac{\pi }{4} \right]$ we get $\displaystyle \frac1{1+\cot^nx}\to0$ and on $\displaystyle \left[ \frac{\pi }{4},\frac{\pi }{3} \right]$ we have $\displaystyle \frac1{1+\cot^nx}\to1,$ hence the Dominated Convergence Theorem applies and $\displaystyle \underset{n\to \infty}{\mathop{\lim }}\,\int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{0\,dx}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{1\,dx}=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}.$
Actually, we can use the equation
$\displaystyle \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{3}}{\lim_{n\rightarrow \infty}\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}$without any doubt.
A theorem says that if $\displaystyle f_{n}(x)$uniformly converges on $\displaystyle [a,b]$, then
$\displaystyle \lim_{n\rightarrow \infty}\int_{a}^{b} {f_{n} (x) \,dx}=\int_{a}^{b} {\lim_{n\rightarrow \infty}f_{n} (x) \,dx}$
Now let $\displaystyle \frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}=f_n (x)$
In this problem, we can easily say that $\displaystyle \forall\epsilon>0, f_n (x)$uniformly converges on$\displaystyle [0,\frac{\pi}{4}-\epsilon],[\frac{\pi}{4}+\epsilon,\frac{\pi}{3}]$,so $\displaystyle \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi}{4}-\epsilon}{ f_{n}(x) \,dx}=\int_{0}^{\frac{\pi}{4}-\epsilon}{\lim_{n\rightarrow \infty} f_{n}(x) \,dx}=0$,$\displaystyle \lim_{n\rightarrow \infty}\int_{\frac{\pi}{4}+\epsilon}^{\frac{\pi}{3 }}{ f_{n}(x) \,dx}=\int_{\frac{\pi}{4}+\epsilon}^{\frac{\pi}{3} }{\lim_{n\rightarrow \infty} f_{n}(x) \,dx}=\frac{\pi}{12}-\epsilon$
so $\displaystyle \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}f_n (x) \ dx=\frac{\pi}{12}-\epsilon+\lim_{n\rightarrow \infty}\int_{\frac{\pi}{4}-\epsilon}^{\frac{\pi }{4}+\epsilon}f_n (x) \ dx$
but $\displaystyle |f_n (x)|<10\forall n,x$,so $\displaystyle |\lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}f_n (x) \ dx-\frac{\pi}{12}|\leq 19\epsilon$
let $\displaystyle \epsilon\rightarrow 0$, we can get $\displaystyle \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}f_n (x) \ dx=\frac{\pi}{12}$!!

8. Originally Posted by NonCommAlg

Solve for real values of $\displaystyle x,y: \ \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt = \frac{\pi x^2}{12}.$
finding all $\displaystyle x,y$ for which the above equality holds requires considering several cases. i'm not going to do that! haha so i will only evaluate the integral.

let $\displaystyle f(t)=\sin^2(xt) \sin^3(yt).$ then:

1) $\displaystyle \lim_{t\to 0} \frac{f(t)}{t^2}=\lim_{t\to\infty} \frac{f(t)}{t^2}=\lim_{t\to 0} \frac{f'(t)}{t}=\lim_{t\to\infty} \frac{f'(t)}{t}=0.$ therefore applying integration by parts twice gives us: $\displaystyle \int_0^{\infty} \frac{f(t)}{t^3} \ dt =\frac{1}{2} \int_0^{\infty} \frac{f''(t)}{t} \ dt.$

2) we know that $\displaystyle \int_0^{\infty} \frac{\sin(ct)}{t} \ dt = \frac{\pi}{2}\text{sgn}(c),$ where $\displaystyle \text{sgn}$ is the sign function.

3) using some routine trigonometry identities we get: $\displaystyle 16f(t)=6\sin(yt) - 2 \sin(3yt) + \sin(3y+2x)t + \sin(3y-2x)t -3\sin(y+2x)t -3 \sin(y-2x)t$.

4) applying 1) to 3) and then using 2) will finally give us:

$\displaystyle \int_0^{\infty} \frac{f(t)}{t^3} \ dt = \frac{\pi}{64}[ 12y^2 \text{sgn}(y) \ - \ (3y+2x)^2 \text{sgn} (3y+2x) \ - \ (3y-2x)^2 \text{sgn}(3y-2x) \ +$

$\displaystyle 3(y+2x)^2 \text{sgn}(y+2x)+3(y-2x)^2\text{sgn}(y-2x) ]. \ \Box$

9. Originally Posted by NonCommAlg
nope! that's not the correct answer! the solution set is the line $\displaystyle y=\frac{2}{3}x.$
But if (x,y) is the solution , then ( -x,y) is also the solution

10. Originally Posted by NonCommAlg
Solve for real values of
Originally Posted by NonCommAlg
nope! that's not the correct answer! the solution set is the line $\displaystyle y=\frac{2}{3}x.$

The problem would be easier to solve if it changes this way:

Solve for real values of
$\displaystyle x,u: \int_0^{\infty} \frac{\sin^2(xt) \sin^3(uxt)}{t^3} \ dt = \frac{\pi x^2}{12}$

or even

Step (1): Show for a fixed $\displaystyle u$, the following integral is proportional to $\displaystyle x^2$, i.e.

$\displaystyle \int_0^{\infty} \frac{\sin^2(xt) \sin^3(uxt)}{t^3} \ dt = f(u)\;x^2$

Step (2): Solve for real values of $\displaystyle u$ such that

$\displaystyle f(u) = \frac{\pi}{12}$

For Step (1), can we use the "differentiation"? hahaha...

11. Originally Posted by simplependulum

But if (x,y) is the solution , then ( -x,y) is also the solution

to moderators: feel free to close this thread whenever you like! cheers to everybody!

12. Originally Posted by simplependulum
luobo : i am looking forward to seeing your solution , actually , I have been shocked to see how you (or someone ) applied magic differentiation in the Integral Bee !! It is really a powerful tool

i think the solution is two straight lines passing throguh the origin (at,bt) and (-at,bt)
simplependulum, the thread starter has put his solution, now it is time to put yours here, of course, not using differentiation! .

13. Originally Posted by NonCommAlg
finding all $\displaystyle x,y$ for which the above equality holds requires considering several cases. i'm not going to do that! haha so i will only evaluate the integral.

let $\displaystyle f(t)=\sin^2(xt) \sin^3(yt).$ then:

1) $\displaystyle \lim_{t\to 0} \frac{f(t)}{t^2}=\lim_{t\to\infty} \frac{f(t)}{t^2}=\lim_{t\to 0} \frac{f'(t)}{t}=\lim_{t\to\infty} \frac{f'(t)}{t}=0.$ therefore applying integration by parts twice gives us: $\displaystyle \int_0^{\infty} \frac{f(t)}{t^3} \ dt =\frac{1}{2} \int_0^{\infty} \frac{f''(t)}{t} \ dt.$

2) we know that $\displaystyle \int_0^{\infty} \frac{\sin(ct)}{t} \ dt = \frac{\pi}{2}\text{sgn}(c),$ where $\displaystyle \text{sgn}$ is the sign function.

3) using some routine trigonometry identities we get: $\displaystyle 16f(t)=6\sin(yt) - 2 \sin(3yt) + \sin(3y+2x)t + \sin(3y-2x)t -3\sin(y+2x)t -3 \sin(y-2x)t$.

4) applying 1) to 3) and then using 2) will finally give us:

$\displaystyle \int_0^{\infty} \frac{f(t)}{t^3} \ dt = \frac{\pi}{64}[ 12y^2 \text{sgn}(y) \ - \ (3y+2x)^2 \text{sgn} (3y+2x) \ - \ (3y-2x)^2 \text{sgn}(3y-2x) \ +$

$\displaystyle 3(y+2x)^2 \text{sgn}(y+2x)+3(y-2x)^2\text{sgn}(y-2x) ]. \ \Box$
I was just overwhelmed by those different cases.

14. Originally Posted by NonCommAlg