Strange integral: int {0,1} [ (1 - t^2)^n * t * sin(xt) ] dt

[tanb56]

Hello everyone! I have a small problem on this integral :
[math]\int_{0}^{1} (1-t^2)^nt \sin(xt) dt[/math]So I started by doing an integration by parts :
[math]u' = t(1-t^2)^n[/math] ---> [math]u = -\frac{(1-t^2)^{n+1}}{2n+2}[/math]v=sin(xt) --> v' = tcos(xt)
Therefore, we have :
[math][\frac{-(1-t^{2})^{n+1}}{2n+2}*sin(xt)]-\int_{0}^{1}{\frac{-(1-t^{2})^{n+1}}{2n+2}*tcos(xt)dt}[/math]The first term is zero, so we have :
[math]\frac{1}{2n+2}*\int_{0}^{1}{-(1-t^{2})^{n+1}tcos(xt)dt}[/math]But from there, I don't really know what to do...
By the way, if you are wondering why I have to solve this integral, it is to show that :
[math]xf'_{x}=(2n+1)f_{n}(x)-f_{n+1}(x)[/math]With : [math]f_{n}(x)=\frac{x^{2n+1}}{2*4*...*(2n)}\int_{0}^{1}{(1-t^{2})^{n}cos(xt)}[/math]

 

[tanb56]

Oops, I made a mistake at the end, it is rather, show that : [imath]xf'_{n}(x)=(2n+1)f_{n}(x)-f_{n+1}(x)[/imath]

 

[Steven G]

If you don't see how to continue, then maybe you did not choose the best substitution.

 

[Steven G]

By the way, if you are wondering why I have to solve this integral, it is to show that :
[math]xf'_{x}=(2n+1)f_{n}(x)-f_{n+1}(x)[/math]With : [math]f_{n}(x)=\frac{x^{2n+1}}{2*4*...*(2n)}\int_{0}^{1}{(1-t^{2})^{n}cos(xt)}[/math]

The last equation you have makes no sense--it's just a minor error.

 

[tanb56]

Oh yes, I had forgotten the dt, thank you! As for my block, I really don't see what variable change I can make. Because if I continue my integration in parts over and over again, I'll always end up with more or less the same thing. And most of all, I don't see what variable change could save me, that's the term [imath]-(1-t^{2})^{n+1}[/imath] that bothers me

 

[khansaheb]

v=sin(xt) --> v' = tcos(xt)

This is incorrect.

The "variable" of integration is 't' - not 'x'. So it should be:

d/dt [sin(xt)] = x * cos(xt) ..................... here 'x' is treated as a constant.

 

[tanb56]

You are absolutely right, sorry for the mistake ^^'. So we have :

[imath]=[\frac{-(1-t^{2})^{n+1}}{2n+2}*sin(xt)]-\int_{0}^{1}{\frac{-(1-t^{2})^{n+1}}{2n+2}}*xcos(xt)dt[/imath]

[imath]=[\frac{-(1-t^{2})^{n+1}}{2n+2}*sin(xt)]-\frac{x}{2n+2}\int_{0}^{1}{-(1-t^{2})^{n+1}cos(xt)dt}[/imath]

And so the first term is always 0, and we have :

[imath]=-\frac{x}{2n+2}\int_{0}^{1}{-(1-t^{2})^{n+1}cos(xt)dt}[/imath]

I may have an idea that would allow me to unblock myself, it would be to pass the cosine in exponential form, but the x bothers me a little.

 

[Steven G]

Oh yes, I had forgotten the dt, thank you! As for my block, I really don't see what variable change I can make. Because if I continue my integration in parts over and over again, I'll always end up with more or less the same thing. And most of all, I don't see what variable change could save me, that's the term [imath]-(1-t^{2})^{n+1}[/imath] that bothers me

Sometimes, when you integrate by parts twice, you get a new integral that is just a multiple of the original integral. You can then solve for the original integral with algebra--no additional integration.

Please try to integral \(\displaystyle \int e^x*sinxdx\) Remember to use integration by parts twice. If you want to show us your work, then please start a new thread.