I suggest directly see "my confusion" part and the last line of my solution to get a quick understand to save your time.
When I was learning the improper integral depending on a parameter chapter in the mathematical analysis lesson, I met this problem which comes from Mathematical Analysis III by Senlin Xu when I was doing its execise.
I almost solve the whole question while there still has some confusion I can't figure out. I don't want to give it up for already spending an afternoon on it.
Question. calculate the integral $$\int_{0}^{+\infty}\frac{\ln(\alpha^{2}+x^{2})}{\beta^{2}+x^{2}}\mathrm{d}x,\quad\beta\neq0.$$Answer is $\frac{\pi}{|\beta|}\ln(|\alpha|+|\beta|)$
My thread is using this rule $\frac{\mathrm{d}}{\mathrm{d}\alpha}(\int_{0}^{+\infty}f(x,\alpha)\mathrm{d}x)=\int_{0}^{+\infty}\frac{\partial f(x,\alpha)}{\partial \alpha}\mathrm{d}x$, and I devided into four step: (1), (2) proof that it satisfies the condtions of exchanging the derivative and integral, (3) find the derivative, (4) integrate.
My counfusion is concentrated on the last line of my solution.
First, why the correct answer has the absolute value sign, which step I ignore to add it?
Second, there is still a integral in last line I can't work out, how to deal with it or it is insoluble which means my solution is somewhere wrong?
My Solution.
(1) First, assume $\alpha \in [a, b] $ and let $$I(\alpha)=\int_{0}^{+\infty}f(x,\alpha)\mathrm{d}x=\int_{0}^{+\infty}\frac{\ln(\alpha^{2}+x^{2})}{\beta^{2}+x^{2}}\mathrm{d}x$$
We have $$\int_{0}^{+\infty}f(x,\alpha)\mathrm{d}x=\int_{0}^{1}f(x,\alpha)\mathrm{d}x+\int_{1}^{+\infty}f(x,\alpha)\mathrm{d}x$$
$\int_{1}^{+\infty}f(x,\alpha)\mathrm{d}x$ is uniformly convergent for $\alpha$. Because $$\int_{1}^{+\infty}f(x,\alpha)\mathrm{d}x = \int_{1}^{+\infty}\frac{\ln(\alpha^{2}+x^{2})}{\beta^{2}+x^{2}}\mathrm{d}x < \int_{1}^{+\infty}\frac{\ln(\max\{|a|,|b|\}^{2}+x^{2})}{\beta^{2}+x^{2}}\mathrm{d}x$$
$\int_{0}^{1}f(x,\alpha)\mathrm{d}x$ is uniformly convergent for $\forall \alpha \in [a, b] \,(0 \notin [a, b])$ and disvergent at $\alpha = 0$, because $$|\int_{0}^{1}f(x,\alpha)\mathrm{d}x |= |\int_{0}^{1}\frac{\ln(\alpha^{2}+x^{2})}{\beta^{2}+x^{2}}\mathrm{d}x| < |\int_{0}^{1}\frac{\ln(\min\{|a|,|b|\}^{2})}{\beta^{2}+x^{2}}\mathrm{d}x|$$ $$|\int_{0}^{1}f(x,0)\mathrm{d}x |= |\int_{0}^{1}\frac{\ln(x^{2})}{\beta^{2}+x^{2}}\mathrm{d}x| > |\int_{0}^{1}\frac{2\ln(x)}{\beta^{2}}\mathrm{d}x|$$ So $I(\alpha)$ is uniformly convergent in any $[a, b] \, (0 \notin [a, b])$.
(2) Find the partia derivative of $f(x, \alpha)$ for $\alpha$ $$\frac{\partial f(x, \alpha)}{\partial\alpha}=\frac\partial{\partial\alpha}(\frac{\ln(\alpha^2+x^2)}{\beta^2+x^2})=\frac{2\alpha}{(\alpha^2+x^2)(\beta^2+x^2)}$$
According to the maximum value theorem, $\exists \alpha_0 \in [a, b]$, make $|\frac{2\alpha}{(\alpha^2+x^2)(\beta^2+x^2)}| \leqslant|\frac{2\alpha_0}{(\alpha_0^2+x^2)(\beta^2+x^2)}|$.
Because
$$\operatorname*{lim}_{x\rightarrow+\infty}\frac{\frac{2\alpha_0}{(\alpha_0^{2}+x^{2})(\beta^{2}+x^{2})}}{\frac{1}{x^{4}}}=1$$
So $\int_{0}^{+\infty}\frac{\partial f(x, \alpha)}{\partial\alpha}\mathrm{d}x=\int_{0}^{+\infty}\frac\partial{\partial\alpha}(\frac{\ln(\alpha^2+x^2)}{\beta^2+x^2})\mathrm{d}x$ is uniformly convergent for $\alpha$.
Also, $f(x, \alpha)$, $\frac{\partial f(x, \alpha)}{\partial\alpha}$ is continuous at $(0, +\infty)\times[a,b]$
(3) For now, assume $\alpha \in [a, b]$, we can use this rule $\frac{\mathrm{d}}{\mathrm{d}\alpha}(\int_{0}^{+\infty}f(x,\alpha)\mathrm{d}x)=\int_{0}^{+\infty}\frac{\partial f(x,\alpha)}{\partial \alpha}\mathrm{d}x$. And we obtain $$while\,\beta\neq\alpha,\,I'(\alpha) =\frac{d}{d\alpha}(\int_{0}^{+\infty}f(x,\alpha)dx) =\int_{0}^{+\infty}\frac{\partial f(x,\alpha)}{\partial\alpha}dx =\frac{2\alpha}{\beta^{2}-\alpha^{2}}\int_{0}^{\infty}(\frac{1}{\alpha^{2}+x^2}-\frac{1}{\beta^{2}+x^2})dx =\frac{2\alpha}{\beta^{2}-\alpha^{2}}(\frac{1}{\alpha}\arctan(\frac{x}{\alpha})-\frac{1}{\beta}\tan(\frac{x}{\beta}))|_{0}^{+\infty} =\frac{\pi}{(\beta+\alpha)\beta}$$ $$while\,\beta=\alpha,\,I'(\beta)=2\beta\int_{0}^{+\infty}\frac{dx}{(\beta^2+x)^2}=2\beta\left.(\frac{(x^2+\beta^2)\arctan\frac x2+\beta x}{2\beta^3x^2+2\beta^5})\right|_{0}^{+\infty}=2\beta\frac{\frac\pi2}{2\beta^3}=\frac\pi{\beta^2} $$ In general, $I'(\alpha)=\frac{\pi}{(\beta+\alpha)\beta}$.
(4) So we can get $$I(\alpha)=\int_{\beta}^{\alpha}\frac{\pi}{(\beta+t)\beta}dt+I(\beta)=\frac{\pi}{\beta}\ln(\alpha+\beta)-\frac{\pi}{\beta}\ln(2\beta)+\int_{0}^{+\infty}\frac{\ln(\beta^{2}+x^{2})}{\beta^{2}+x^{2}}dx$$