Wikipedia:Reference desk/Archives/Mathematics/2009 September 12

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September 12

Simplify integral

Can this be simplified? ${\frac {d}{dv}}\int _{-\infty }^{\infty }\phi _{1}(v_{1})\left(\int _{-v_{1}}^{v-v_{1}}\phi _{2}(v_{2})dv_{2}\right)dv_{1}$

Probably some bad choice of variables here, I'll rewrite it first.

${\frac {d}{dv}}\int _{-\infty }^{\infty }f(y)\left(\int _{-y}^{v-y}g(x)dx\right)dy$

Thanks in advance--Yanwen (talk) 03:54, 12 September 2009 (UTC)

I get f*g where * means convolution. I'm kinda rusty with this type of problem so take for what it's worth.--RDBury (talk) 05:05, 12 September 2009 (UTC)

This appears to be an exercise (I'm guessing homework) in using the Leibniz integral rule and the first fundamental theorem of calculus. 67.122.211.205 (talk) 19:09, 12 September 2009 (UTC)

Ahha. Leibniz integral rule was what I was missing.

${\frac {d}{dv}}\int _{-\infty }^{\infty }f(y)\left(\int _{-y}^{v-y}g(x)dx\right)dy$

$\int _{-\infty }^{\infty }{\frac {\partial }{\partial v}}f(y)\left(G(v-y)-G(-y)\right)dy$

$\int _{-\infty }^{\infty }f(y)g(v-y)dy$--Yanwen (talk) 01:39, 13 September 2009 (UTC)

what is the value of k that must be added to 7,16,43,79 so that they are in proportion?

what is the value of k that must be added to 7,16,43,79 so that they are in proportion? —Preceding unsigned comment added by 122.168.68.145 (talk) 09:32, 12 September 2009 (UTC)

5.--RDBury (talk) 14:16, 12 September 2009 (UTC)

only if 16=19...Tinfoilcat (talk) 16:19, 12 September 2009 (UTC)

In proportion to what? --Tango (talk) 16:20, 12 September 2009 (UTC)

My interpretation was that you're supposed to add a constant k to each term and end up with a bit of a geometric series (i.e. al, al^2, al^3, al^4). This problem does not have a solution Tinfoilcat (talk) 16:28, 12 September 2009 (UTC)

Perhaps they don't have to be consecutive terms from a geometric series? Or could it just be that one must divide the next, which divides the next, etc.? --Tango (talk) 16:41, 12 September 2009 (UTC)

The OP probably means that the first and second number should be in the same ratio as the third and fourth. In this case, RDBury's answer is correct. -- Meni Rosenfeld (talk) 19:49, 12 September 2009 (UTC)

As Meni stated, RDBury's answer is correct BUT, doesn't this strike anyone as a HW problem? hydnjo (talk) 21:08, 12 September 2009 (UTC)

Fourier Transforms

Resolved

Could anyone let me know whether I'm doing this wrong?

For $f(x)=\left\{{\begin{array}{lr}cos(x):|x|<{\frac {\pi }{2}}\\0\,\,\,\,\,\,\,\,\,\,\,\,\,:|x|>{\frac {\pi }{2}}\end{array}}\right.$, do we have ${\tilde {f}}(k)=\int _{\frac {-\pi }{2}}^{\frac {\pi }{2}}f(x)e^{-ikx}\,dx$ (that's the definition I'm using, so I know that bit's correct) $=Re(\int _{\frac {-\pi }{2}}^{\frac {\pi }{2}}e^{ix}e^{-ikx}\,dx)=Re[{\frac {e^{(1-k)ix}}{(1-k)i}}]_{\frac {-\pi }{2}}^{\frac {\pi }{2}}=(*)\,Re[{\frac {-i\cdot 2i}{1-k}}\,sin(1-k)({\frac {\pi }{2}})]=2{\frac {sin((1-k)({\frac {\pi }{2}}))}{1-k}}$. Is that correct?

Because I thought ${\tilde {f'}}(k)=ik{\tilde {f}}(x)$, but then $f'(x)=-sin(x)$ so ${\tilde {f'}}(k)={\tilde {-sin(x)}}={\frac {-1}{i}}Im[(\int _{\frac {-\pi }{2}}^{\frac {\pi }{2}}e^{ix}e^{-ikx}\,dx]$ which unless I'm being stupid becomes 0 at $(*)$, doesn't it? But then that doesn't equal $ik\cdot 2{\frac {sin((1-k)({\frac {\pi }{2}}))}{1-k}}$, does it? Perhaps I'm being dense.

Following that, I'm trying to show 'by considering the F.T. of the above function and its derivative', that $\int _{0}^{\infty }{\frac {{\frac {\pi ^{2}}{4}}-cos^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}=\int _{0}^{\infty }{\frac {t^{2}-cos^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}={\frac {\pi }{4}}$, but I haven't a clue how to do it! I know it's related to Parseval's relation, but other than possibly moving the 2 integrals together and canceling a $({\frac {\pi ^{2}}{4}}-t^{2})$, then showing it's equal to 0 (which still doesn't solve the problem fully), i'm not sure how to move along, or whether that's even a good idea. Could anyone suggest anything perhaps?

Thanks a lot,

Delaypoems101 (talk) 23:50, 12 September 2009 (UTC)

Let me point out one error you made in computing the Fourier Transform of $f(x)$. It is true that $\cos(x)=Re\{e^{ix}\}$; however

$\int \cos(x)e^{-ikx}dx=\int Re\{e^{ix}\}\;e^{-ikx}dx\neq Re\{\int e^{ix}\;e^{-ikx}dx\}$.

Instead use the relation $2\cos(x)=e^{ix}+e^{-ix}$ to compute the Fourier transform integral (Hint: ${\tilde {f}}(k)$ is sum of two shifted sinc functions). Abecedare (talk) 00:11, 13 September 2009 (UTC)

Yeah, we can caluculate the Fourier transform directly using integration by parts twice.

${\hat {f}}(\xi )=\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\!\!\cos(x)e^{-ix\xi }\ dx={\frac {2\cos \left({\frac {\pi }{2}}\xi \right)}{1-\xi ^{2}}}$

As for the first integral, well, you have some problems. For example, the integrand has an even-ordered pole when $t=\pm {\frac {\pi }{2}}$. This singularity is non-removable since the numerator of the integrand is always positive for real t. In fact

$\int _{0}^{y}{\frac {{\frac {\pi ^{2}}{4}}-\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt\to \infty \ {\mbox{as}}\ y\to \infty \ .$

As for the second integral, well, again:

$\int _{0}^{y}{\frac {t^{2}-\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt\to \infty \ {\mbox{as}}\ y\to \infty \ .$ ~~ Dr Dec (Talk) ~~ 01:19, 13 September 2009 (UTC)

Oh god, how stupid of me, there's no minus in the numerator - sorry about wasting your time! Do things work better with $\int _{0}^{y}{\frac {{\frac {\pi ^{2}}{4}}\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt$ and $\int _{0}^{y}{\frac {t^{2}\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt$?Delaypoems101 (talk) 02:08, 13 September 2009 (UTC)

No matter - got it sorted: thanks all! Delaypoems101 (talk) 03:41, 13 September 2009 (UTC)