solving an integral-derivative problem

Forum for the GRE subject test in mathematics.
GREMATH
Posts: 6
Joined: Mon Jan 03, 2011 11:17 pm

solving an integral-derivative problem

Postby GREMATH » Thu Mar 10, 2011 7:08 pm

Please help me with solving this derivative:

d/dx (∫(cos(xt)/t)dt)

the integral is from x to 0.

thanks,

bobzilla21
Posts: 3
Joined: Thu Mar 10, 2011 8:47 pm

Re: solving an integral-derivative problem

Postby bobzilla21 » Thu Mar 10, 2011 8:55 pm

I thought calculus isn't on the GRE?

congvan
Posts: 9
Joined: Sat Mar 06, 2010 4:44 am

Re: solving an integral-derivative problem

Postby congvan » Thu Mar 10, 2011 10:11 pm

bobzilla21 wrote:I thought calculus isn't on the GRE?


Not on the GRE General but it is 50% of the GRE Math subject test

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: solving an integral-derivative problem

Postby blitzer6266 » Fri Mar 11, 2011 4:50 am

There might be an easier way to do this problem, but the only obvious way I would approach it is using taylor expansion. We know \cos(xt) = \sum_0^\infty{(-1)^n(xt)^{2n} \over (2n)! so {d \over dx} \int_x^0{\sum_0^\infty{(-1)^n(xt)^{2n} \over t(2n)!} dt = {d \over dx} \int_x^0 {1 \over t} dt + {d \over dx} \int_x^0{\sum_1^\infty{(-1)^n(xt)^{2n} \over t(2n)!} dt = -{1 \over x} -{d \over dx} \sum_1^\infty{{(-1)^n(x)^{2n}(x)^{2n} \over (2n)(2n)! }  = -{1 \over x} -2 \sum_1^\infty{(-1)^nx^{4n-1} \over (2n)! = {1 \over x} -2 \sum_0^\infty{(-1)^nx^{4n-1} \over (2n)!} ={ 1 - 2\cos(x^2) \over x}
I glossed over a lot of the details, but i think this is correct. Feel free to make corrections

PNT
Posts: 37
Joined: Fri Mar 11, 2011 9:01 pm

Re: solving an integral-derivative problem

Postby PNT » Fri Mar 11, 2011 9:36 pm

There is an easier way then using taylor series. You can see that when you differentiate first you "lose" d(the variable limit)/dx * (f(the variable limit).
So \frac{d}{dx}\int_{x}^{0}\frac{\cos(xt)}{t}dt=-\frac{\cos(x^{2})}{x}-\int_{x}^{0}\sin(xt)dt  = \frac{1-2\cos(x^{2})}{x}

which is just the Leibniz integral rule

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: solving an integral-derivative problem

Postby blitzer6266 » Fri Mar 11, 2011 10:40 pm

Looks good. I was thinking about using Leibniz, but I didn't think I could given that the integrand isn't continuous at t=0. How do you justify this?

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: solving an integral-derivative problem

Postby owlpride » Fri Mar 11, 2011 11:28 pm

Edit: I didn't see that PNT had already referenced the Leibnitz integral rule, but I'll keep it posted for the benefit of future readers because it's useful to know for the GRE:

\frac{d}{dx}\int_{g(x)}^{h(x)} f(x,t)dt = f(x,h(x)) \frac{dh}{dx} - f(x,g(x)) \frac{dg}{dx} + \int_{g(x)}^{h(x)} {\frac{d}{dx}f(x,t) dt}

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: solving an integral-derivative problem

Postby blitzer6266 » Sat Mar 12, 2011 1:26 am

Actually I'm going to change my answer to does not exist. For example, if x=1, the integral of cos(t)/t from 0 to 1 does not converge so it doesn't make sense to define the derivative at that point. One way to fix this problem would be to change the question to "what is

\lim_{h \to 0^+} {d \over dx} \int_x^h {\cos(xt) \over t} dt for x > 0 (and a similar definition for x < 0 )

(Note: This is not the same as the original question)




Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”



Who is online

Users browsing this forum: Bing [Bot], itosformula and 3 guests