# Differentiating under the integral sign

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
Mentor
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?

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)##.

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
Mentor
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
Mentor
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.

Mr Davis 97