Integration by parts SinIntegral[x]

[eclayj]

Homework Statement

Calculate the following integral exactly (no approximations) by the method of integration by parts:

∫₀^t SinIntegral[x] dx

Homework Equations

the following hints are given:

D[SinIntegral[x], x] = Sinc[x]; and
SinIntegral[0] = 0

The Attempt at a Solution

The integration by parts forrmula is ∫_a^bu[x]v^'[x]dx = u[x]v[x]|_a^b-∫_a^bv[x]u^'[x]dx.

The problem I have is that whether I let u[x]=1 and v'[x]= SinIntegral[x], or u[x]=SinIntegral[x] and v'[x]= 1, either way the integral on the right is more complex than what I started with (either ∫₀^txSinc[x]; or ∫₀^tSinIntegral[x].

Any help is appreciated. I hope all the programming tags work here, otherwise this will look like a mess. It may still look like a mess

So it seems to me that, although the question requires the differentiation by parts method, such method does not work because it you get more complicated integrals

 

[eclayj]

Wait, maybe I undertstand it. Do you have to use integration by parts method twice? i.e., letting u[x] = SinIntegral[x], and v^\[Prime][x] = 1; u^\[Prime][x] = Sinc[x] ; v[x] = x
Which after applying the formula would get you to tSinIntegral[t] - ∫₀^tSinc[x]dx.

Then you run the integration by parts once again to find n = ∫₀^tSinc[x]dx.


Then the final answer will be tSinIntegral[t] - n?

This seems a little tedious, but the question specifically asks you to use the integration by parts formula, so I am guessing this must be it, unless I am missing something obvious... which is usually the case.

 

[eclayj]

nope. Still stuck

 

[SammyS]

Do you know what Sinc(x)․x is ?

 

[eclayj]

Thanks Sammy. I had never encoutered the Sinc Function before. After your hint, and a little help from google, I found that Sinc[x]= (Sin[x])/x. So xSinc[x] should be Sin[x], which makes the problem much easier. I think that should have been another hint given, I'm only working in Calc II ;).

 

[HACR]

Is it sin int(t)*t+cos(t)-1? since D(sin int(x))=sinc(x)=sin(x)/x, u=sin int(x). If you look a int^udv=uv-int^v*du, du is on the RHS.