Differentiating under the integral sign

[Mr Davis 97]

I am looking at a solution to an integral using differentiation under the integral sign. So let $\displaystyle f(t) = \frac{\log (tx+1)}{x^2+1}$. Then, through calculation, $\displaystyle f'(t) = \frac{\pi t + 2 \log (2) - 4 \log (t+1)}{4(1+t^2)}$. The solution immediately goes to say that $\displaystyle f(t) = \frac{\log (2) \arctan (t)}{2} + \frac{\pi \log (t^2+1)}{8} - \int_0^t \frac{\log (t+1)}{t^2+1} ~ dt$. Could someone make this step a bit clearer? How does the LHS just become $f(t)$?

 

[Mark44]

I am looking at a solution to an integral using differentiation under the integral sign. So let $\displaystyle f(t) = \frac{\log (tx+1)}{x^2+1}$. Then, through calculation, $\displaystyle f'(t) = \frac{\pi t + 2 \log (2) - 4 \log (t+1)}{4(1+t^2)}$. The solution immediately goes to say that $\displaystyle f(t) = \frac{\log (2) \arctan (t)}{2} + \frac{\pi \log (t^2+1)}{8} - \int_0^t \frac{\log (t+1)}{t^2+1} ~ dt$. Could someone make this step a bit clearer? How does the LHS just become $f(t)$?

The antiderivative of the derivative of a function is the function (plus an arbitrary constant). It's as simple as that.

In symbols: $\int f'(t) dt = f(t) + C$

As a definite integral, along the lines of the equation you showed,
$\int_0^t f'(x)dx = f(t) - f(0)$. Here -f(0) is the constant C I showed above.

 

[Stephen Tashi]

I am looking at a solution to an integral

What is the integral whose solution you seek?

 

[Mr Davis 97]

What is the integral whose solution you seek?

Sorry, I made a mistake. The first line should be $\displaystyle f(t) = \int_0^1 \frac{\log (tx+1)}{x^2+1} ~ dx$. The integral I seek is $f(1)$.

 

[Mr Davis 97]

The antiderivative of the derivative of a function is the function (plus an arbitrary constant). It's as simple as that.

In symbols: $\int f'(t) dt = f(t) + C$

As a definite integral, along the lines of the equation you showed,
$\int_0^t f'(x)dx = f(t) - f(0)$. Here -f(0) is the constant C I showed above.

I guess the use of $t$ is a limit of integration is confusing me, since t is already a variable elsewhere. For example, say that $\frac{dy}{dx} = 2x$. Then $\int_0^x dy = \int_0^x 2x ~dx$. Doesn't this latter statement seem kind of weird? Why is x a limit of integration if it is already used as a variable? Also, what does $\int_0^x dy$ turn out to look like?

 

[Mark44]

I guess the use of $t$ is a limit of integration is confusing me, since t is already a variable elsewhere. For example, say that $\frac{dy}{dx} = 2x$. Then $\int_0^x dy = \int_0^x 2x ~dx$. Doesn't this latter statement seem kind of weird? Why is x a limit of integration if it is already used as a variable? Also, what does $\int_0^x dy$ turn out to look like?

A limit of integration shouldn't be used also as a variable in the integrand (AKA "dummy variable"). You should note that in my 2nd example I was careful to use different variables for the two different roles.

If $\frac{dy}{dx} = 2x$, then $dy = 2x dx$, so $\int_0^t dy = \int_0^t 2x dx \Rightarrow y(t) - y(0) = x^2|_0^t \Rightarrow y(t) = t^2 + y(0)$

 

[Mr Davis 97]

A limit of integration shouldn't be used also as a variable in the integrand (AKA "dummy variable"). You should note that in my 2nd example I was careful to use different variables for the two different roles.

If $\frac{dy}{dx} = 2x$, then $dy = 2x dx$, so $\int_0^t dy = \int_0^t 2x dx \Rightarrow y(t) - y(0) = x^2|_0^t \Rightarrow y(t) = t^2 + y(0)$

Okay, so the fact the solution I am reading does use the same variable as the variable in the integrand is not technically wrong, but is just bad "style"?

 

[Mark44]

Okay, so the fact the solution I am reading does use the same variable as the variable in the integrand is not technically wrong, but is just bad "style"?

Yes, and confusing to some.