Integral of log and exponential with power

Question

This integral sounds quite complex and I could not find an approximate equivalent. Any hopes for solving: $$\int_{0}^{+\infty} x\log(1+x^2)\,e^{- B x}\,dx$$

Answer 1

For starters, substitute $x^2\mapsto x,~$ as follows :

$$\begin{align}I ~&=~\quad\int_0^\infty~x~\ln(1+x^2)~e^{\large-Bx}~dx ~=~\frac12\quad\int_0^\infty\ln(1~+~x^2)~e^{\large-B\sqrt{x^2}}~d(x^2)~=~ \\\\ ~&=~\frac12~\int_0^\infty\quad\ln(1+x)~e^{\large-B\sqrt x}~dx ~=~\frac12~\bigg[\int_0^\infty\ln(1+Ax)~e^{\large-B\sqrt x}~dx\bigg]_{A=1} ~=~\frac{J(1)}2 \end{align}$$

Now, employ differentiation under the integral sign :

$$\begin{align}J'(A) ~&=~\int_0^\infty\frac x{1+Ax}~e^{\large-B\sqrt x}~dx ~=~\frac{d^2}{dB^2}~\int_0^\infty\frac1{1+Ax}~e^{\large-B\sqrt x}~dx~=~ \\\\ ~&=~\frac{d^2}{dB^2}~\bigg\{\frac1A~\bigg[\sin\frac B{\sqrt A}\bigg(\pi-2\text{ Si }\frac B{\sqrt A}\bigg)-2~\cos\frac B{\sqrt A}\text{ Ci }\dfrac B{\sqrt A}\bigg]\bigg\}, \end{align}$$

which, when integrated with regard to A, yields the rather beautiful expression :

$$\frac{J(A)}2~=~\frac{d^2}{dB^2}\bigg[\text{ Ci}^2~\dfrac B{\sqrt A}+\text{ Si}^2~\dfrac B{\sqrt A}-\pi\text{ Si }\dfrac B{\sqrt A}\bigg],$$

which, when evaluated at $A=1,~$ gives $~I~=~\dfrac{J(1)}2~=~\dfrac{d^2}{dB^2}~\bigg[\text{ Ci}^2~B+\text{ Si}^2~B-\pi\text{ Si }B\bigg],$

coinciding with the result provided by Mariusz Iwaniuk in the comments. QED.