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My question is regarding integration. I have a complex function that needs to be integrated and its a definite integral. The thing is when I use Wolfram Alpha to integrate this function it gives me nothing i.e its unable to compute it. However if I remove the boundaries of integration i.e I make my integral an indefinite integral, Wolfram Alpha is able to compute. Now my question is Can I take the result I obtained for the indefinite integral and just evaluate for the boundary limits to evaluate my definite integral ?

If my analysis is correct, then why wouldn't Wolfram alpha give the result anyways?

using Wolfram Alpha, if I try 

integrate(exp(-v)/(1+sv^-1))

then I get the following result

-e^(-v)-e^s s Ei(-s-v)

While if I try

integrate(exp(-v)/(1+sv^-1),{v,1,+infinity})

I get nothing!

George
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  • I could do it by inputing "integral | exp(-v)/(1+sv^-1) | v = 1 to infinity" but it exceeds standard computational time – Nathan Wride Oct 28 '14 at 04:17
  • As long as there are no discontinuities you can evaluate using the boundary limits. Think of trying this with a function that isn't continuous and see what you can make it do. If your s is real and greater than zero then I believe you can do this because your integrand is continuous between 1 and infinity. – Bill Oct 28 '14 at 04:29
  • thanks bill. In general how would I check if my function has a discontinuity or not? – George Oct 28 '14 at 04:42
  • yes this is what I mean – George Oct 28 '14 at 04:51
  • @George: I suspect that to be completely general, you can't check for every possible discontinuity. But checking that you have no denominators that can be zero would be a start. Try plotting your integrand with s= -1, s=0 and s=4 and see what those look like. – Bill Oct 28 '14 at 05:58
  • thanks Bill that is right. Actually my s is real and non negative so I think I am fine. – George Oct 28 '14 at 14:27

2 Answers2

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since you tagged this Mathematica:

by specifying an appropriate assumption on s we get the expected result:

 Integrate[Exp[-v]/(1 + s/v) , {v, 1, Infinity},  Assumptions -> {s > -1}]

 --> 1/E + E^s s ExpIntegralEi[-1 - s]

I don't know if alpha has some similar syntax to add assumptions..

additionally if we try a finite integral:

 Integrate[Exp[-v]/(1 + s/v) , {v, 1, 2} ]

mathematica returns a conditional expression that tells us the result is valid for s>-1 or s<-2. For some reason it doesn't give such result for the infinite case however.

agentp
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Yes, you can take the result obtained for the indefinite integral and use to calculate the definite integral. When I try to run your request at Wolfram Alpha, here's what I get:

Wolfram Alpha response to asked definite integration

As you can see in the highlighted portion at the bottom left of the above picture, Wolfram Alpha didn't complete your request because it exceeded the standard computation time. This is because they need to offer some extra features for Wolfram Alpha Pro users to pay for the service. One of this features is extended computation time.

Wolfram Alpha is a business, and this is one of the ways it makes money. See for yourself, it'll offer you the pro service if you click the "Try again with additional computational time" on the bottom right.

If you just break down the definite integration between first the indefinite integral (which it can handle) and then calculate the boundary values and take the difference, it seems to work fine:

Indefinite integration

Calculation between boundaries

This is mathematically correct because that is how definite integrals are calculated.

However your input has an sv in the dividend. Wolfram Alpha is taking it to mean s*v, which might not be what you meant—if sv is a variable on it's own, I suggest you rename it to s or something else. The point is that if s is indeed a variable, if you take a look at the plot in the answer, there seems to be a ridge due to the -∞ term, so for some values of s that ridge might be within your integration curve, and then the integral can't be calculated, as Bill pointed out in his comment to your question.

kadu
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  • this is exactly what i obtained, and this is what i meant by my question. I guess I am looking for an argument that convinces me why i can take the indefinite integration result and compute it at the limits to obtain my definite integral. I think that if this is the case, than wolfram alpha would have done it this way – George Oct 28 '14 at 04:28
  • thanks i understand what you mean. Thank you for taking time to explain. what do you exactly mean by break down the indefinite integral? Do you mean evaluate it at the limits? – George Oct 28 '14 at 04:52
  • Oh, my mistake, I meant to write break down the definite integration between the indefinite integration (which it can handle) and the calculation at the boundaries (which it can also handle). I'll edit the answer. I'll add another point at the end addressing [Bill's comment](http://stackoverflow.com/questions/26600664/wolfram-alpha-is-able-to-integrate-an-indefinite-integral-but-not-a-definite-int/26600849?noredirect=1#comment41815733_26600664) because there's one point that seems unclear to me, so stay tuned. – kadu Oct 28 '14 at 04:55
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    I think I know what I should do. You have referenced in your answer the following [that is how definite integrals are calculated](http://en.wikipedia.org/wiki/Integral#Introduction), after reading this page, I noticed there is a section called (Second fundamental theorem of calculus) please check it out. It says I have to prove that my function is integrable over [a b] so that if I use the result of the indefinite integral it is correct. – George Oct 28 '14 at 05:08
  • This is compatible with what Bill suggested above. Do you happen to know how I can prove that this function is integrable? – George Oct 28 '14 at 05:13
  • A function is integrable if it doesn't have any discontinuities. A point where it tends to `∞` or `-∞` is one type of [discontinuity](http://en.wikipedia.org/wiki/Classification_of_discontinuities), and it *seems* to be the only type you need worry about. *If `s` is indeed a separate variable*, *for some values of `s` there will be a discontinuity in your integration interval*. From this point onwards, it becomes about mathematics and the particulars of your problem, and outside of the scope of this site. You can use wolfram to set values to `s` and see what happens. – kadu Oct 28 '14 at 05:23