feynman's technique or somethin

[sharp prawn]

so we have the integral $$ I(a) = \int_{0}^{a} \frac{\ln (ax + 1)}{x^2 + 1} \mathrm{d} x$$
How do we differentiate this with respect to a?

[forest ridge]

fijokazż

[vestal ice]

why do you have the bound with a as well?

haven't seen feynman's technique used with these but it's the same principle

here you use leibniz rule for differentiating an integral

[vestal ice]

$$
\begin{aligned}
I'(a)&=\frac{d}{da} \left[ \int_{u(a)}^{v(a)} f(a,x) dx \right] \
&= \frac{d}{da}(f(a,v(a)) - f(a,u(a))) + \int_{u(a)}^{v(a)} \frac{ \partial }{ \partial a } f(a,x)dx\
&= f(a,v(a))v'(a) -f(a,u(a))u'(a) + \int_{u(a)}^{v(a)} \frac{ \partial }{ \partial a } f(a,x)dx
\end{aligned}
$$

[forest ridge]

artemetra

[vestal ice]

in your case:
$$
\begin{aligned}
I'(a) &= \frac{\ln(a^2+1)}{a^2+1}\left(\frac{d}{da} a\right) - 0 + \int_{0}^{a} \frac{ \partial }{ \partial a } \frac{\ln(ax+1)}{a^2+1} dx\
&= \frac{\ln(a^2+1)}{a^2+1} +\int_{0}^{a} \frac{ \partial }{ \partial a } \frac{\ln(ax+1)}{a^2+1} dx\
&= \frac{\ln(a^2+1)}{a^2+1} + \int_0^a \frac{x}{(ax+1)(x^2+1)} dx
\end{aligned}
$$

[forest ridge]

artemetra

[vestal ice]

@sharp prawn

[sharp prawn]

interesting. wouldnt we need to multiply them? why is it addition?

[vestal ice]

that's the general form for leibniz rule when your integral bounds have the variable you are differentiating w.r.t.

[sharp prawn]

i didnt know that. i always thought we used the chain rule. thanks very much, that cleaned some things up

[vestal ice]

again i haven't seen feynman's technique used with integrals where the new variable is also in the bound so idk how useful this is for that but yeah this is the general form for differentiating the integral sign

[vestal ice]

it is just the chain rule really

[sharp prawn]

.solved