solving an integralderivative problem
solving an integralderivative problem
Please help me with solving this derivative:
d/dx (∫(cos(xt)/t)dt)
the integral is from x to 0.
thanks,
d/dx (∫(cos(xt)/t)dt)
the integral is from x to 0.
thanks,

 Posts: 3
 Joined: Thu Mar 10, 2011 8:47 pm
Re: solving an integralderivative problem
I thought calculus isn't on the GRE?
Re: solving an integralderivative problem
Not on the GRE General but it is 50% of the GRE Math subject testbobzilla21 wrote:I thought calculus isn't on the GRE?

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: solving an integralderivative 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 $${d \over dx} \int_x^0{\sum_0^\infty{(1)^n(xt)^{2n} \over t(2n)!} dt = {d \over dx} \int_x^0 {1 \over t} dt + {d \over dx} \int_x^0{\sum_1^\infty{(1)^n(xt)^{2n} \over t(2n)!} dt$$ $$= {1 \over x} {d \over dx} \sum_1^\infty{{(1)^n(x)^{2n}(x)^{2n} \over (2n)(2n)! } = {1 \over x} 2 \sum_1^\infty{(1)^nx^{4n1} \over (2n)!$$ $$= {1 \over x} 2 \sum_0^\infty{(1)^nx^{4n1} \over (2n)!} ={ 1  2\cos(x^2) \over x}$$
I glossed over a lot of the details, but i think this is correct. Feel free to make corrections
I glossed over a lot of the details, but i think this is correct. Feel free to make corrections
Re: solving an integralderivative 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 $$\frac{d}{dx}\int_{x}^{0}\frac{\cos(xt)}{t}dt=\frac{\cos(x^{2})}{x}\int_{x}^{0}\sin(xt)dt = \frac{12\cos(x^{2})}{x}$$
which is just the Leibniz integral rule
So $$\frac{d}{dx}\int_{x}^{0}\frac{\cos(xt)}{t}dt=\frac{\cos(x^{2})}{x}\int_{x}^{0}\sin(xt)dt = \frac{12\cos(x^{2})}{x}$$
which is just the Leibniz integral rule

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: solving an integralderivative 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?
Re: solving an integralderivative 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:
$$\frac{d}{dx}\int_{g(x)}^{h(x)} f(x,t)dt = f(x,h(x)) \frac{dh}{dx}  f(x,g(x)) \frac{dg}{dx} + \int_{g(x)}^{h(x)} {\frac{d}{dx}f(x,t) dt}$$
$$\frac{d}{dx}\int_{g(x)}^{h(x)} f(x,t)dt = f(x,h(x)) \frac{dh}{dx}  f(x,g(x)) \frac{dg}{dx} + \int_{g(x)}^{h(x)} {\frac{d}{dx}f(x,t) dt}$$

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: solving an integralderivative 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)
$$\lim_{h \to 0^+} {d \over dx} \int_x^h {\cos(xt) \over t} dt$$ for $$x > 0$$ (and a similar definition for $$x < 0$$ )
(Note: This is not the same as the original question)