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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 The answer is from @Quantum spaghettification (who from his profile stats seems to be no longer active on this site):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 $$ I(1,0)$$

Community
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Srini
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1 Answers1

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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.

Zaid Alyafeai
  • 14,343