🚀 Add to Chrome – It’s Free - YouTube Summarizer
Category: N/A
No summary available.
00:00
here's a nice integral and by nice I mean it seems hostile at first glance but it turns out to have a really elegant solution development and the solution development here involves my favorite integration technique I'm talking about feynman's approach of
00:16
differentiating under the integral sign so we're going to call our integral I and we're going to Define an integral function I of some parameter a and where exactly do we place this parameter well a sensible Choice here would be as part
00:33
of the argument of the sine function so we have e to the negative x squared times the sine of a x squared divided by x squared DX and the reason I'm calling it sensible is because if you differentiate partially with respect to
00:49
a the sine of ax squared you get the derivative of the sine function as the cosine function so you have cosine ax squared and because of the the chain rule the derivative of a with respect to a is one and x squared is just a
01:07
constant multiple because we're taking the partial derivative so we have this extra factor of x squared which will cancel out with the denominator quite nicely so yeah this is a good choice so as per the approach we're going to
01:23
differentiate the integral function with respect to the parameter a and the golden question here is can we perform the switch up of the integration and the differentiation operators well if you look at the integrand well it consists of a gaussian term here e to
01:41
the negative x squared and we have a bounded sine function and all of this is being divided by x squared which is an increasing function on this interval so we're dividing by an increasing function or we can say that we're multiplying by
01:56
the decreasing Function One by x squared so yeah on this interval there are no problems regarding convergence or boundedness so yes indeed we can perform the switch up and because of the switch up the total derivative with respect to
02:12
a becomes a partial one so we're differentiating partially with respect to a e to the negative x squared times the sine of a x squared divided by x squared and we're carrying out the integration with respect to X
02:29
because we're differentiating partially with respect to a the X terms here are constants so we have the integral from 0 to Infinity of e to the negative x squared divided by x squared times the derivative of the sine function which is the cosine function so we have cosine ax
02:46
squared and because because of the chain rule we need the derivative of the argument which sorts out in this case to x squared and this multiple of x squared cancels out quite nicely with our denominator here so this implies the
03:02
derivative of I with respect to a is the integral from 0 to Infinity of e to the negative x squared times the cosine of ax squared DX next up we're going to make use of Euler's wonderful formula whereby we
03:18
know that e to the i x equals the cosine of x plus I times the sine of x and notice that our inner Grand contains the cosine term so we're only interested in the real part of this complex
03:34
exponential so if we need cosine a x squared then this means that we need the real part of e to the I a x squared Okay cool so all of this implies that the derivative of I with respect to a is the
03:50
real part of the integral from 0 to Infinity of e to the negative x squared times e to the I a x squared DX so multiplying out these two exponentials you're left with the real part of the integral from 0 to Infinity of e to the
04:06
I'm going to factor out a negative x squared term here so that leaves me with 1 minus I times a and what exactly is the structure that you've that you've finally obtained after all
04:23
of this awesome mathematics well you'll be quite happy to know that this integral is just an example of a generalized a sort of generalized gaussian integral where if you have the integral from 0 to Infinity of e to the
04:39
negative s x squared DX where s is a complex number with a non-negative real part and in this case we do have a non-negative real part we have a real part equal to one so anyway this
04:55
integral sorts out to one half of the square root of Pi by S so that means we have a pretty nice structure for I prime we have I prime of a being equal to this factor of one-half then this factor
05:12
of square root pi times the real part of the reciprocal of the square root of your complex number which in this case is just 1 minus I times a and we can write this as uh
05:28
half the square root of pi times the real part of 1 minus I times a to the negative one-half and now that we have the derivative of I with respect to a completely in terms of the parameter a
05:44
alone we can proceed to recover our integral function I from its derivative by integrating with respect to a so on the right hand side you have square root Pi by 2 times the real part of the
06:01
integral of 1 minus I times a to the negative one-half d a so this implies that on the left hand side you have I of a being equal to the square root of Pi by 2 times the real
06:18
part of 1 minus I times a to the negative one half plus one is just one-half right and downstairs again you have this one half factor and you have to divide by the derivative of 1 minus I times a with respect to a right which is
06:35
just a negative I the imaginary unit Okay cool so we're dividing by negative I and we have this constant of integration that we're going to deal with later as well okay so finally we have I of a being
06:51
equal to square root Pi by two and wait a second these factors cancel out quite nicely once again so we have square root pi times the real part of one minus I times a the square root of that divided by
07:07
negative I now the good news is we know that uh 1 by I equals negative I which is just awesome I mean I think this this little equation here is just incredibly beautiful and
07:24
another one of the many many reasons to love complex analysis so we know this so we can write negative I the reciprocal of negative I as just plain old I so we need the real part of I times the square root of 1 minus I times a
07:41
and we have this constant of integration as well so how exactly do we figure out the value of the constant of integration well recall that your integral function I of a was defined as the integral from 0 to Infinity of e to the negative x
07:58
squared times the sine of a x squared divided by x squared so if you plug in a equal to zero then that means that upstairs you have sine of 0 which is zero so the entire integrand collapses
08:16
to zero giving you I of zero being equal to zero so that's a pretty useful piece of information so using a equals zero we have zero equal to the square root of pi times the real part of I times the
08:31
square root of one minus zero which is one anyway plus the constant of integration that we're looking for and the real part of I is zero anyway so this implies that your constant of integration is conveniently zero Okay
08:46
cool so that means we can finally turn our attention to our Target case and our Target case is that of a being equal to one so this implies that your target integral which is I of one equals the
09:04
square root of pi times the real part of I times the square root of 1 minus I times a and a in our Target case is just one so we're now interested in the real part of this complex number which is quite easy to extract all we have to do
09:23
is use the polar representation of complex numbers so let Z be the complex number one minus I and in the polar form for that you need the modulus of Z which in this case is the square root of 2 and
09:39
the argument of Z and remember we're only interested in the principal Branch so here we have the inverse tangent of the image imaginary part which is negative one divided by the real part which is one which sorts out to negative pi by four so that means in the polar
09:54
form you have Z equal to 1 minus I equal to the square root of 2 times e to the negative I times pi by 4. and we're interested in the square root of 1 minus
10:09
I and the square root is just an exponent of uh one half right so this implies that the square root of 1 minus I equals the square root of the square root of 2 times e to the negative I Pi by 8 now
10:25
and again we're interested in multiplying the square root term by the imaginary unit I so carrying out this multiplication and once again expanding the complex exponential using Euler's wonderful formula we have the square
10:41
root of we have I times the square root of the square root of 2. times the cosine of negative pi by eight is the same as the cosine of pi by 8 because cosine is an even function on the contrary the sine is an odd function so
10:56
the negative sign just pops out so you have minus I times sine Pi by 8. and multiplying out this I term here that gives you I times the cosine term and I squared
11:13
times the sine term now negative I squared is just positive one okay cool so finally we have I times the square root of 1 minus I and we need the real part of this and that is pretty clear that that is
11:30
the square root of the square root of 2 times the sine of Pi by 8. okay nice this is awesome so all we needed was this real part times the square root of Pi and that gives you your target integral I so finally we can
11:48
write here that the integral from 0 to Infinity of e to the negative x squared times sine X squared divided by x squared DX equals the square root of pi times the square root of 2 times the
12:03
sine of Pi by 8. a really nice result indeed I hope you enjoyed the video be sure to like And subscribe thank you see you next time