I think this involves Leibniz theorem

[Raghav Gupta]

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?

 

[mooncrater]

Hi!
I think this question doesn't require the leibnitz rule. I think it CAN be done by :
f(x)=∫₁^x(logt/1+t)dt
Substituting t=1/k
dt=-dk/k²
1 will be 1
And x will become 1/x
Then f(x)=∫₁^1/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.

 

[Raghav Gupta]

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?

 

[mooncrater]

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).

 

[Raghav Gupta]

I am then getting,
same result.
How you have evaluated the integral?

 

[mooncrater]

I am saying that:
f(x)+f(1/x)=
∫₁^x(log t/1+t)dt[i.e the original integrand]+∫₁^xlog[log t /t(t+1)]dt [i have replaced k with t]

 

[Raghav Gupta]

I am saying that:
f(x)+f(1/x)=
∫₁^x(log t/1+t)dt[i.e the original integrand]+∫₁^xlog[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,
∫₁^x (logt/t)
= (logt)²/2
Putting limits,
1/2 (logx)²
Got it
Thanks buddy.
Can you tell what was wrong in my attempt?

 

[mooncrater]

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...

 

[Raghav Gupta]

But we are getting that answer only if we don't take into account constant of integration, in post 1 attempt.

 

[Dick]

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 $x=1$.

 

[Raghav Gupta]

We'll get 2f(1) = 0 + c
Now f(1) = 0
So c= 0.
How did you think that we should put x = 1 ?

 

[Dick]

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?

 

[Raghav Gupta]

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.

 

[Dick]

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?

 

[Raghav Gupta]

Yeah. Thanks.

 

[Raghav Gupta]

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?

 

[Dick]

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 $C$.

 

[mooncrater]

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

Shouldn't $f'(1/x) =-xlog x/1+x$

 

[Raghav Gupta]

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

 

[Ray Vickson]

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 $-f'(1/x)/x^2$.

 

[Raghav Gupta]

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?

 

[Ray Vickson]

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.

 

[Raghav Gupta]

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
$(d/dx)f(x) + (d/dx)f(1/x)] = \ln(x)/x$

 

[Ray Vickson]

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 $-\frac{1}{x^2} f'(x)$.