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Indefinite Integration

  • Thread starter Saitama
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  • #1
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Homework Statement


The value of [tex]\int_0^{1} (\prod_{r=1}^{n} (x+r))(\sum_{k=1}^{n} \frac{1}{x+k}) dx[/tex] equals:
a)n
b)n!
c)(n+1)!
d)n.n!

(Can someone tell me how to make bigger parentheses using latex?)

Homework Equations





The Attempt at a Solution


I know that the question becomes a lot easier if i put n=1 or 2 and then integrate. But i was wondering if there is any proper way to solve it. I can't go further after expanding the given expression.
[tex]\int_{0}^{1} (x+1)(x+2)......(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+.....\frac{1}{x+n})dx[/tex]
which is equal to
[tex]\int_{0}^{1} \sum_{r=1}^n \frac{(x+n)!}{x+r}[/tex]
I am stuck now, i can't find any way further.

Any help is appreciated. :smile:
 

Answers and Replies

  • #2
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I still don't have any clue.
 
  • #3
ehild
Homework Helper
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What is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ?

ehild
 
  • #4
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What is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ?

ehild
Derivative of (x+1)(x+2)=2x+3
Derivative of (x+1)(x+2)(x+3)=3x2+12x+11

I still don't get any idea how this would help?
 
  • #5
ehild
Homework Helper
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Do not simplify.

d/dx[(x+1)(x+2)]=(x+2)+(x+1)=(x+1)(x+2)/(x+1)+(x+1)(x+2)/(x+2)=[(x+1)(x+2)][1/(x+1)+1/(x+2)]

d/dx[(x+1)(x+2)(x+3)]=(x+2)(x+3)+(x+1)(x+3)+(x+1)(x+2)=[(x+1)(x+2)(x+3)][1/(x+1)+1/(x+2)+1(x+3)]

Do you see?

ehild
 
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  • #6
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Wait, i guess i have got it, i will be back in a few hours with a solution.
Thanks for the help! :smile:

EDIT: Oops, seems like you posted just a few seconds before me.
 
  • #7
ehild
Homework Helper
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Wait, i guess i have got it, i will be back in a few hours with a solution.
Thanks for the help! :smile:

EDIT: Oops, seems like you posted just a few seconds before me.
Well, I looking forward to the solution:smile:

ehild
 
  • #8
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So here's the solution:
Let
[tex]t=(x+1)(x+2)(x+3).......(x+n)[/tex]
[tex]\ln t=\ln(x+1)+\ln(x+2)+\ln(x+3)......+\ln(x+n)[/tex]
[tex]\frac{1}{t}\frac{dt}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}......\frac{1}{x+n}[/tex]
[tex]dt=(x+1)(x+2)(x+3)....(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}........+\frac{1}{x+n})dx[/tex]
Therefore the original question becomes ∫dt, upper limit is (n+1).n! and lower limit is n!.
Solving, i get the answer as d) option.

Thanks for the help! :smile:
 
  • #9
Mentallic
Homework Helper
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By the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that.

I myself use /left( and /right) which looks like: [tex]\left(\frac{1}{1}\right)[/tex] but there are other more specific "fonts" you can use, which ranges from smallest to largest:

/big( [tex]\big(\frac{1}{1}\big)[/tex]
/Big( [tex]\Big(\frac{1}{1}\Big)[/tex]
/bigg( [tex]\bigg(\frac{1}{1}\bigg)[/tex]
/Bigg( [tex]\Bigg(\frac{1}{1}\Bigg)[/tex]

And there could be others, but that should about cover it.
 
  • #10
ehild
Homework Helper
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1,854
So here's the solution:
Let
[tex]t=(x+1)(x+2)(x+3).......(x+n)[/tex]
[tex]\ln t=\ln(x+1)+\ln(x+2)+\ln(x+3)......+\ln(x+n)[/tex]

[tex]\frac{1}{t}\frac{dt}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}......\frac{1}{x+n}[/tex]
[tex]dt=(x+1)(x+2)(x+3)....(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}........+\frac{1}{x+n})dx[/tex]
That is ingenious! You are really cool:cool:

Therefore the original question becomes ∫dt, upper limit is (n+1).n! and lower limit is n!.
Solving, i get the answer as d) option.

Thanks for the help! :smile:
Very good. You are better and better every day!

ehild
 
  • #11
3,812
92
By the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that.

I myself use /left( and /right) which looks like: [tex]\left(\frac{1}{1}\right)[/tex] but there are other more specific "fonts" you can use, which ranges from smallest to largest:

/big( [tex]\big(\frac{1}{1}\big)[/tex]
/Big( [tex]\Big(\frac{1}{1}\Big)[/tex]
/bigg( [tex]\bigg(\frac{1}{1}\bigg)[/tex]
/Bigg( [tex]\Bigg(\frac{1}{1}\Bigg)[/tex]

And there could be others, but that should about cover it.
Thanks Mentallic, really helpful. :smile:
That is ingenious! You are really cool:cool:
I did nothing, you almost solved it by giving me a simple hint. Thanks! :smile:
 
  • #12
ehild
Homework Helper
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1,854
That method of using logarithm was entirely your idea :cool:. I did not think of it.

ehild
 

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