$$ \sqrt{p}-\sqrt{\sqrt{1+p^2}-1} $$ My attempt： This function have four branch points $$ p=0,p= i,p=-i,p=\infty $$
and defined $$ -\pi <\mathrm{arg}\left( p \right) <\pi \,\,,-\frac{\pi}{2}<\mathrm{arg}\left( p+i \right) <\frac{3\pi}{2},-\frac{3\pi}{2}<\mathrm{arg}\left( p-i \right) <\frac{\pi}{2} $$ enter image description here $$ l_1:\sqrt{1+p^2}-1=\left| 1+p^2 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left\{ \mathrm{arg}\left( p+i \right) +\mathrm{arg}\left( p-i \right) \right\}}-1=i\sqrt{y^2-1}-1;\sqrt{\sqrt{1+p^2}-1}=\left| i\sqrt{y^2-1}-1 \right|^{\frac{1}{2}}e^{\frac{i}{2}\left( -\mathrm{arc}\tan \sqrt{y^2-1}+\pi \right)}=\sqrt{y}ie^{-i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}} \\ \int_{\infty}^1{\left( e^{i\frac{\pi}{4}}\sqrt{y}-ie^{-i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}}\sqrt{y} \right)}e^{iyt}i\mathrm{d}y $$ $$ l_2:\sqrt{1+p^2}-1=\left| 1+p^2 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left\{ \mathrm{arg}\left( p+i \right) +\mathrm{arg}\left( p-i \right) \right\}}-1=-i\sqrt{y^2-1}-1;\sqrt{\sqrt{1+p^2}-1}=\left| -i\sqrt{y^2-1}-1 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left( \mathrm{arc}\tan \sqrt{y^2-1}-\pi \right)}=-i\sqrt{y}e^{i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}} \\ \int_1^{\infty}{\left( e^{i\frac{\pi}{4}}\sqrt{y}+i\sqrt{y}e^{i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}} \right)}e^{iyt}i\mathrm{d}y $$ $$ l_3:\sqrt{1+p^2}-1=\left| 1+p^2 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left\{ \mathrm{arg}\left( p+i \right) +\mathrm{arg}\left( p-i \right) \right\}}-1=\sqrt{y^2+1}-1;\sqrt{\sqrt{1+p^2}-1}=\sqrt{\sqrt{y^2+1}-1} \\ \int_{\infty}^0{-\left( i\sqrt{y}-\sqrt{\sqrt{y^2+1}-1} \right) e^{-yt}\mathrm{d}y}=\int_0^{\infty}{\left( i\sqrt{y}-\sqrt{\sqrt{1+y^2}-1} \right) e^{-yt}\mathrm{d}y} $$ $$ l_4:\sqrt{1+p^2}-1=\left| 1+p^2 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left\{ \mathrm{arg}\left( p+i \right) +\mathrm{arg}\left( p-i \right) \right\}}-1=\sqrt{y^2+1}-1;\sqrt{\sqrt{1+p^2}-1}=\sqrt{\sqrt{y^2+1}-1} \\ \int_0^{\infty}{-\left( -i\sqrt{y}-\sqrt{\sqrt{1+y^2}-1} \right) e^{-yt}\mathrm{d}y}=\int_0^{\infty}{\left( i\sqrt{y}+\sqrt{\sqrt{1+y^2}-1} \right) e^{-yt}\mathrm{d}y} $$ $$ l_5:\sqrt{1+p^2}-1=\left| 1+p^2 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left\{ \mathrm{arg}\left( p+i \right) +\mathrm{arg}\left( p-i \right) \right\}}-1=i\sqrt{y^2-1}-1;\sqrt{\sqrt{1+p^2}-1}=\left| i\sqrt{y^2-1}-1 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left( -\mathrm{arc}\tan \sqrt{y^2-1}+\pi \right)}=\sqrt{y}ie^{-i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}} \\ \int_{\infty}^1{-i\left( e^{-i\frac{\pi}{4}}\sqrt{y}-\sqrt{y}ie^{-i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}} \right) e^{-iyt}\mathrm{d}y} $$ $$ l_6:\sqrt{1+p^2}-1=\left| 1+p^2 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left\{ \mathrm{arg}\left( p+i \right) +\mathrm{arg}\left( p-i \right) \right\}}-1=-i\sqrt{y^2-1}-1;\sqrt{\sqrt{1+p^2}-1}=\left| -i\sqrt{y^2-1}-1 \right|^{\frac{1}{2}}e^{i\frac{1}{2}\left( \mathrm{arc}\tan \sqrt{y^2-1}-\pi \right)}=-i\sqrt{y}e^{i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}} \\ \int_1^{\infty}{-i\left( e^{-i\frac{\pi}{4}}\sqrt{y}+i\sqrt{y}e^{i\frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1}} \right) e^{-iyt}\mathrm{d}y} $$ and integrals along the circle tend to zero，by Cauchy's theorem we have $$ \int_{\gamma -i\infty}^{\gamma +i\infty}{\left( \sqrt{p}-\sqrt{\sqrt{1+p^2}-1} \right) e^{pt}\mathrm{d}p}+2i\int_0^{\infty}{\sqrt{y}e^{-yt}\mathrm{d}y}-2\int_1^{\infty}{\cos \left( \frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1} \right) \sqrt{y}\left( e^{iyt}-e^{-iyt} \right) \mathrm{d}y}=0 $$ This means $$ \mathscr{L} ^{-1}\left\{ \sqrt{p}-\sqrt{\sqrt{1+p^2}-1} \right\} =\frac{1}{2\pi i}\int_{\gamma -i\infty}^{\gamma +i\infty}{\left( \sqrt{p}-\sqrt{\sqrt{1+p^2}-1} \right) e^{pt}\mathrm{d}p}=\frac{2}{\pi}\int_1^{\infty}{\cos \left( \frac{1}{2}\mathrm{arc}\tan \sqrt{y^2-1} \right) \sqrt{y}\sin \left( yt \right) \mathrm{d}y}-\frac{1}{\pi}\int_0^{\infty}{\sqrt{y}e^{-yt}\mathrm{d}y} \\ =\frac{\sqrt{2}}{\pi}\int_1^{\infty}{\sqrt{y+1}\sin \left( yt \right) \mathrm{d}y}-\frac{1}{\pi}\int_0^{\infty}{\sqrt{y}e^{-yt}\mathrm{d}y} $$ but the The first one on the right is divergent，what's wrong with that? Can someone help me，or use other methods to calculate this inversion.Any suggestions would be appreciated.
