# Feynman integration - help with this old answer

I need help understanding an answer: https://math.stackexchange.com/a/1808872/335418 . The answer is from @Quantum spaghettification (who from his profile stats seems to be no longer active on this site):

I asked for clarification there, but it doesn't look like he/she is going to see it. Can someone help me understand this?

Snippet from that answer that I don't understand:

Now integrating w.r.t. $a$ gives us: $$\frac{\partial I(t,0)}{\partial t}=-\int^0_{\infty}\frac{t}{a^2+t^2}da$$ making the substitution $a=t \tan(\theta)$ this becomes: $$\frac{\partial I(t,0)}{\partial t}=-\int^0_{\pi/2} da$$ $$=\frac{\pi}{2}$$ Then integrating w.r.t $t$: $$I(1,0)=\int^1_0 \frac{\pi}{2} dt$$ $$=\frac{\pi}{2}$$

My questions:

1. When the RHS of the first equation is begin integrate between $\infty$ and $0$, from the previous step (not pasted here), the LHS is equivalent to $$\int\limits_{\infty}^{0}\frac{\partial^2 I(t,a)}{\partial a\partial t} da$$ right? How is this the same as $$\frac{\partial I(t,0)}{\partial t}$$

2. Similarly how is $$\int\limits_{0}^{1} \frac{\partial I(t,0)}{\partial t} dt$$ same as $$I(1,0)$$

• Re: 2, integration and differentiation are inverse operations, no? – J. M. is a poor mathematician Dec 28 '16 at 5:33
• Agree, but I was expecting #2 to be something like $I(1,0) - I(0,0)$. i.e. Subtraction of the limits for $t$ keeping '$a$' constant. – Srini Dec 28 '16 at 6:09
• This is because $\lim_{a\to+\infty}\frac{\partial I(t,a)}{\partial t} = 0$. For large $a$, $I(t,a) \sim O(a^{-1})$ because of the $e^{-at}$ factor in the integrand.. – achille hui Dec 28 '16 at 6:40

Note by the Fundamental theorem of calculus

$$\int^b_a f(x) \,dx= F(b)-F(a)$$

Where $F'(x) = f(x)$.

Assuming that

$$I(t,a)=\int^\infty_{0}\frac{e^{-ax}\sin^2(xt)}{x^2}dx$$

Then

$$\int\limits_{0}^{1} \frac{\partial I(t,0)}{\partial t} dt = I(1,0)-I(0,0)$$

We conclude that

$$I(0,0) = \lim_{t \to 0}\int^\infty_{0}\frac{\sin^2(xt)}{x^2}dx = 0$$

Similarly

$$\int\limits_{\infty}^{0}\frac{\partial^2 I(t,a)}{\partial a\partial t} da = \frac{\partial I(t,0)}{\partial a}-\frac{\partial I(t,\infty)}{\partial a}$$

Where

$$\frac{\partial I(t,\infty)}{\partial a} = -\lim_{a\to \infty}\int^\infty_{0}\frac{e^{-ax}\sin^2(xt)}{x}dx = 0$$

There are here lots of assumptions that we can switch the limit and the integrals.

• Thanks for explaining each step explicitly. – Srini Dec 28 '16 at 13:33
• @Srini, ur welcome. It is a really important property that I used a lot. – Zaid Alyafeai Dec 29 '16 at 7:27