## 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

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

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

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

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

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

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

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

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)

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