I think this involves Leibniz theorem

In summary: D_{1/x} f(1/x) = f'(1/x)##.In summary, for x>0, the function f(x) is defined as the integral from 1 to x of (log t)/(1+t). When f(x) is added to f(1/x), the result is equal to (1/2)(log x)^2. This can be found by substituting t=1/k and solving for f(1/x). The constant of integration is found by setting x=1 and solving for f(1). The derivative of f(1/x) is not equal to f'(1/x) and can be denoted as D_x f(1/x) = -f'(1/x)/
  • #1
Raghav Gupta
1,011
76

Homework Statement


For x> 0 , let f(x) = $$\int _1^x \frac{log t dt }{1+t}$$
Then ## f(x) + f(1/x)## is equal to :

A. ##¼ (log x)^2 ##

B. ## ½ (log x)^2 ##

C. ##log x ##

D. ## ¼ log x^2 ##

Homework Equations


Suppose f(x) = $$\int _1^x g(t)$$
Then by Leibniz rule ,
f' (x) = g(x)

The Attempt at a Solution


I found f' (x) = logx/(1+x)
f' ( 1/x) = logx/x(x+1)
f' (x) + f'(1/x) = logx/x
Now what to do?
We can integrate both sides but a constant will appear which we don't know?
 
Physics news on Phys.org
  • #2
Hi!
I think this question doesn't require the leibnitz rule. I think it CAN be done by :
f(x)=∫1x(logt/1+t)dt
Substituting t=1/k
dt=-dk/k2
1 will be 1
And x will become 1/x
Then f(x)=∫11/x[log k/k(k+1)]dk
Then you can find f(1/x) from this equation and add that to the original f(x).
But I am getting an answer that is not any of the options . I think I am getting a wrong f(1/x). If this method works , can you give the steps? I would be very thankful.
 
  • #3
I am getting solving your attempt,
f(x) + f(1/x) = $$ \int_1^{1/x} \frac{logk dk}{k(k+1)} + \int_1^{x} \frac{logk dk}{k(k+1)} $$
How we'll solve further?
 
  • #4
Raghav Gupta said:
I am getting solving your attempt,
f(x) + f(1/x) = $$ \int_1^{1/x} \frac{logk dk}{k(k+1)} + \int_1^{x} \frac{logk dk}{k(k+1)} $$
How we'll solve further?
Instead of using the new f(x) , use the original one( given in the question).
 
  • #5
I am then getting,
same result.
How you have evaluated the integral?
 
  • #6
I am saying that:
f(x)+f(1/x)=
1x(log t/1+t)dt[i.e the original integrand]+∫1xlog[log t /t(t+1)]dt [i have replaced k with t]
 
  • #7
mooncrater said:
I am saying that:
f(x)+f(1/x)=
1x(log t/1+t)dt[i.e the original integrand]+∫1xlog[log t /t(t+1)]dt [i have replaced k with t]
Mistake in bold part.
That bold part should not be there.
Then we'll simply get,
1x (logt/t)
= (logt)2/2
Putting limits,
1/2 (logx)2
Got it
Thanks buddy.
Can you tell what was wrong in my attempt?
 
  • Like
Likes mooncrater
  • #8
Are you talking about post #1? No ., sorry I don't know how to do this question by that method. You are right about the unavaibility of constant of integration that would be needed...
 
  • #9
But we are getting that answer only if we don't take into account constant of integration, in post 1 attempt.
 
  • #10
Raghav Gupta said:
But we are getting that answer only if we don't take into account constant of integration, in post 1 attempt.

You can determine the constant of integration. Put ##x=1##.
 
  • Like
Likes Raghav Gupta
  • #11
We'll get 2f(1) = 0 + c
Now f(1) = 0
So c= 0.
How did you think that we should put x = 1 ?
 
  • #12
Raghav Gupta said:
We'll get 2f(1) = 0 + c
Now f(1) = 0
So c= 0.
How did you think that we should put x = 1 ?

So ##f(1)=0## tells you the constant of integration must be zero. What's the question?
 
  • #13
But instead of writing question again, I think you could see it yourself in post 1.
I think you mean something else.
Integral from 1 to 1 is zero.
 
  • #14
Raghav Gupta said:
But instead of writing question again, I think you could see it yourself in post 1.
I think you mean something else.
Integral from 1 to 1 is zero.

Integrating both sides as in post 1 gives you ##\frac{1}{2} (log(x))^2+C=f(x)+f(1/x)## where ##C## is the constant of integration. Putting ##x=1## shows that ##C=0##. Isn't that what you wanted to know?
 
  • #15
Yeah. Thanks.
 
  • #16
Dick said:
Integrating both sides as in post 1 gives you ##\frac{1}{2} (log(x))^2+C=f(x)+f(1/x)## where ##C## is the constant of integration. Putting ##x=1## shows that ##C=0##. Isn't that what you wanted to know?
Yes, but How you got the idea that we'll find C if we'll put x = 1?
 
  • #17
Raghav Gupta said:
Yes, but How you got the idea that we'll find C if we'll put x = 1?

It's the same way you eliminate a constant of integration when you are solving other differential equations. You substitute a point where the functions of ##x## are known and find ##C##.
 
Last edited:
  • Like
Likes Raghav Gupta
  • #18
Raghav Gupta said:
f' ( 1/x) = logx/x(x+1)
Shouldn't ##f'(1/x) =-xlog x/1+x ##
 
  • #19
mooncrater said:
Shouldn't ##f'(1/x) =-xlog x/1+x ##

Why it should be?
Apply Leibniz rule. Can you show what you have done?
For x> 0 , let f(x) = $$\int _1^x \frac{log t dt }{1+t}$$
Then ## f(x) + f(1/x)## is equal to :Suppose f(x) = $$\int _1^x g(t)$$
Then by Leibniz rule ,
f' (x) = g(x)

The Attempt at a Solution


I found f' (x) = logx/(1+x)
f' ( 1/x) = logx/x(x+1)
f' (x) + f'(1/x) = logx/x
 
  • #20
Raghav Gupta said:
Why it should be?
Apply Leibniz rule. Can you show what you have done?
For x> 0 , let f(x) = $$\int _1^x \frac{log t dt }{1+t}$$
Then ## f(x) + f(1/x)## is equal to :Suppose f(x) = $$\int _1^x g(t)$$
Then by Leibniz rule ,
f' (x) = g(x)

The Attempt at a Solution


I found f' (x) = logx/(1+x)
f' ( 1/x) = logx/x(x+1)
f' (x) + f'(1/x) = logx/x

The notation ##f'(1/x)## should be avoided in this problem; it means ##\left. f'(t) \right|_{t = 1/x}## in standard mathematical notation. That is very different from ##(d/dx) f(1/x)##, which is equal to ##-f'(1/x)/x^2##.
 
  • #21
Ray Vickson said:
The notation ##f'(1/x)## should be avoided in this problem; it means ##\left. f'(t) \right|_{t = 1/x}## in standard mathematical notation.
But I wanted to imply that only.
Then I think I am correct?
 
  • #22
Raghav Gupta said:
But I wanted to imply that only.
Then I think I am correct?

The correct result is ##(d/dx)[f(x) + f(1/x)] = \ln(x)/x## and that is not what you wrote.
 
  • #23
Ray Vickson said:
The correct result is ##(d/dx)[f(x) + f(1/x)] = \ln(x)/x## and that is not what you wrote.
Okay, got it. I wanted to imply this
##(d/dx)f(x) + (d/dx)f(1/x)] = \ln(x)/x##
 
  • #24
Raghav Gupta said:
Okay, got it. I wanted to imply this
##(d/dx)f(x) + (d/dx)f(1/x)] = \ln(x)/x##

This is correct, but ##(d/dx) f(1/x) \neq f'(1/x)##; in fact, ##(d/dx) f(1/x) = - f'(1/x)/x^2##.

Notation is always an issue in problems of this type, so sometimes the best compromise is to use a "differential operator" such as ##D_x## to denote ##d/dx##, so that the ##x##-derivative of ##f(1/x)## could be denoted as ##D_x f(1/x)##, which equals ##-\frac{1}{x^2} f'(x)##.
 
  • Like
Likes Raghav Gupta

What is Leibniz theorem?

Leibniz theorem, also known as the generalized binomial theorem, is a mathematical theorem that provides a formula for expanding powers of binomials to any real or complex exponent. It is named after the German mathematician Gottfried Wilhelm Leibniz.

What is the significance of Leibniz theorem?

Leibniz theorem is significant because it allows us to easily calculate the coefficients of a binomial expansion without having to manually multiply out each term. It also has many applications in probability, statistics, and calculus.

What is the formula for Leibniz theorem?

The formula for Leibniz theorem is (x + y)^n = ∑(n choose k) * x^(n-k) * y^k, where n is the exponent and k ranges from 0 to n. (n choose k) represents the binomial coefficient, which can be calculated using n! / (k! * (n-k)!).

Can Leibniz theorem be applied to negative exponents?

Yes, Leibniz theorem can be applied to negative exponents, as long as the base of the binomial is non-zero and the exponent is a real number. In this case, the formula would be (x + y)^-n = 1 / ∑(n choose k) * x^(n-k) * y^k.

How is Leibniz theorem related to the binomial distribution?

Leibniz theorem is closely related to the binomial distribution, which is a probability distribution that describes the number of successes in a fixed number of independent trials. The binomial distribution can be calculated using the binomial formula, which is a special case of Leibniz theorem.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
836
  • Calculus and Beyond Homework Help
Replies
4
Views
613
  • Calculus and Beyond Homework Help
Replies
3
Views
495
  • Calculus and Beyond Homework Help
Replies
3
Views
128
  • Calculus and Beyond Homework Help
Replies
12
Views
917
  • Calculus and Beyond Homework Help
Replies
1
Views
614
  • Calculus and Beyond Homework Help
Replies
5
Views
536
  • Calculus and Beyond Homework Help
Replies
8
Views
301
  • Calculus and Beyond Homework Help
Replies
15
Views
735
  • Calculus and Beyond Homework Help
Replies
23
Views
833
Back
Top