The integral $\int\ln(x)\cos(1+(\ln(x))^2)\,dx$ Help with a integral calculus please!?
The equation is 
$$\int\ln(x)\cos(1+(\ln(x))^2)\,dx$$
My teacher told me, i have to use substitution? but i can't still solve it.
I've been solving this last week but still i can't get the answer, please help me guys. Thanks!
 A: First, consider the following integral:
$$I(t)=\int\sin(1+(t+\ln(x))^2)\ dx$$
By letting $x=e^u$ and $\sin(\theta)=\Im(e^{i\theta})$, we get
$$I(t)=\int e^u\sin(1+(t+u)^2)\ du=\Im\int e^{u+(1+(t+u)^2)i}\ du$$
This may then be solving using the error function,
$$\int e^{u+(1+(t+u)^2)i}\ du=\frac{-\sqrt{\pi i}e^{i+\frac{(2it+1)^2}{4i}}\operatorname{erf}\left(\frac{2iu+2it+1}2\sqrt i\right)}2$$
It then follows that
$$I'(t)=\frac d{dt}\int\sin(1+(t+\ln(x))^2)\ dx=\int2(t+\ln(x))\cos(1+(t+\ln(x))^2)\ dx\\I'(t)=\Im\frac d{dt}\frac{-\sqrt{\pi i}e^{i+\frac{(2it+1)^2}{4i}}\operatorname{erf}\left(\frac{2iu+2it+1}2\sqrt i\right)}2$$
Thus,
$$\frac12I'(0)=\int\ln(x)\cos(1+(\ln(x))^2)\ dx$$
Evaluating the derivative, one gets
$$\begin{align}\frac12I'(t)&=\Im\frac{-\sqrt{\pi i}}4\frac d{dt}e^{i+\frac{(2it+1)^2}{4i}}\operatorname{erf}\left(\frac{2iu+2it+1}2\sqrt i\right)\\&=\Im\frac{-\sqrt{\pi i}}4e^{i+\frac{(2it+1)^2}{4i}}\left[(2it+1)\operatorname{erf}\left(\frac{2iu+2it+1}2\sqrt i\right)+\frac{4i}{\sqrt\pi}e^{-\left(\frac{2iu+2it+1}2\sqrt i\right)^2}\right]\end{align}$$
Finally,

$$\int\ln(x)\cos(1+(\ln(x))^2)\ dx=\Im\frac{-\sqrt{\pi i}}4\left[e^{i+\frac1{4i}}\left[\operatorname{erf}\left(\frac{2iu+1}2\sqrt i\right)+\frac{4i}{\sqrt\pi}e^{-\left(\frac{2iu+1}2\sqrt i\right)^2}\right]\right]$$

Where $u=\ln(x)$.
A: First, substitute via $u=1+\ln^2(x)$ and use the complex exponential form of cosine to get
\begin{align*}
\int \ln(x) \cos(1+\ln^2(x)) \,dx 
&= \int \sqrt{u-1}\, \cos(u)\, \frac{\exp(\sqrt{u-1})}{2\sqrt{u-1}} \,du \\
&= \frac{1}{2} \int \cos(u)\,\exp(\sqrt{u-1}) \,du\\
&= \frac{1}{2} \int  \frac{1}{2}[\exp(iu)+\exp(-iu)] \,\exp(\sqrt{u-1}) \,du \\
&= \frac{1}{4}
\left[ {\int \exp(iu)  \,\exp(\sqrt{u-1}) \,du}
+
 \int  \exp(-iu) \,\exp(\sqrt{u-1}) \,du
\right]\\
&= \frac{1}{4}
\left[ I_1 + I_2 \right]
\end{align*}
Let's deal with $I_1$ first.
\begin{align*}
I_1 
&= {\int \exp(iu)  \,\exp(\sqrt{u-1}) \,du}\\
&= \underbrace{-\exp(\sqrt{u-1}) i \exp(iu) }_{\gamma_1}
- \int -i\exp(iu)\frac{\exp(\sqrt{u-1})}{2\sqrt{u-1}} \,du\\
&= -\gamma_1 + ie^i\int \exp(a^2i+a) \,da\\
&= -\gamma_1 + ie^i\int \exp\left( i\left[a-\frac{i}{2}\right]^2 +\frac{i}{4} \right) \,da\\
&= -\gamma_1 + i\exp\left(\frac{5i}{4}\right)\int 
\exp\left( \left[a\sqrt{i} - \frac{i^{3/2}}{2} \right]^2 \right) \,da\\
&= -\gamma_1 + \sqrt{i}\exp\left(\frac{5i}{4}\right)\int 
\exp\left( z^2 \right) \,da\\
&= -\gamma_1 + \frac{\sqrt{i\pi}}{2}\exp\left(\frac{5i}{4}\right)\,\text{erfi}(z) 
\end{align*}
where the first step uses integration by parts, then the substitution 
$a=\sqrt{u-1}$ is used, followed by the substitution $z=a\sqrt{i} - i^{3/2}/2$.
Recall that $\text{erfi}(z)=-i\,\text{erf}(iz)$.
Now for $I_2$, using a similar strategy.
\begin{align*}
I_2
&= {\int \exp(-iu)  \,\exp(\sqrt{u-1}) \,du}\\
&= \underbrace{\exp(\sqrt{u-1}) i \exp(-iu)}_{\gamma_2}
 - \int i\exp(-iu)\frac{\exp(\sqrt{u-1})}{2\sqrt{u-1}} \,du\\
&= \gamma_2 - 
i\int \exp(-ia^2-i+a) \,da\\
&= \gamma_2 - 
ie^{-i}\int \exp\left( -i\left[a + \frac{i}{2}\right]^2 -\frac{i}{4} \right) \,da\\
&= \gamma_2 - 
i \exp\left( \frac{-5i}{4} \right) 
\int \exp\left( -\left[ a\sqrt{i} + \frac{i^{3/2}}{2} \right]^2 \right) \,da\\
&= \gamma_2 - 
\sqrt{i} \exp\left( \frac{-5i}{4} \right) 
\int \exp\left( -\zeta^2 \right) \,da\\
&= \gamma_2 - 
\frac{\sqrt{i\pi}}{2} 
\exp\left( \frac{-5i}{4} \right) \,
\text{erf}(\zeta)\\
\end{align*}
using the same subsitution with $a$ and with $\zeta = a\sqrt{i} + i^{3/2}/2$.
Ok, now let's simplify the terms without the error functions:
\begin{align*}
-\gamma_1 + \gamma_2 &=
-\exp(\sqrt{u-1}) i \exp(iu) +
\exp(\sqrt{u-1}) i \exp(-iu) \\
&=
-\underbrace{\exp(\sqrt{u-1})}_{x}i[\underbrace{e^{iu} - e^{-iu}}_{2i\sin(u)}] \\
&=
2x\sin(1+\ln^2(x)) \\
\end{align*}
where we used the complex exponential form of sine.
Next we need to be able to undo the substitutions:
\begin{align*}
z 
&= \sqrt{i} a - \frac{i^{3/2}}{2}\\
&= \sqrt{i} \sqrt{u-1} - \frac{i^{3/2}}{2}\\
&= \sqrt{i} \ln(x) - \frac{i^{3/2}}{2}\\
&= \frac{1}{2} \sqrt{i}\left[ 2\ln(x) - i \right]\\
\zeta 
&= \sqrt{i} a + \frac{i^{3/2}}{2}\\
&= \sqrt{i} \sqrt{u-1} + \frac{i^{3/2}}{2}\\
&= \sqrt{i} \ln(x) + \frac{i^{3/2}}{2}\\
&= \frac{1}{2} \sqrt{i}\left[ 2\ln(x) + i \right]
\end{align*}
Now we can put it all together:
\begin{align*}
\int & \ln(x)  \,\cos(1+\ln^2(x)) \,dx \\
&= \frac{1}{4} \left[ I_1 + I_2 \right]\\
&= \frac{1}{4} \left[
-\gamma_1 + \frac{\sqrt{i\pi}}{2}\exp\left(\frac{5i}{4}\right)\,\text{erfi}(z) 
+
\gamma_2 - 
\frac{\sqrt{i\pi}}{2} 
\exp\left( \frac{-5i}{4} \right) 
\text{erf}(\zeta)
\right]\\
&=
\frac{1}{4} \left(
-\gamma_1 + \gamma_2 + 
\frac{\sqrt{i\pi}}{2}
\exp\left(\frac{-5i}{4}\right)
\left[
\exp\left( \frac{5i}{2} \right)
\text{erfi}(z) 
 -  
\text{erf}(\zeta)\right]
\right)\\
&=
\frac{1}{4} \left(
2x\sin(1+\ln^2(x))  + 
\frac{\sqrt{i\pi}}{2}
\exp\left(\frac{-5i}{4}\right)
\left[
\exp\left( \frac{5i}{2} \right)
\text{erfi}(z) 
 -  
\text{erf}(\zeta)\right]
\right)\\
&=
\frac{1}{8} \left(
4x\sin(1+\ln^2(x))  + 
{\sqrt{i\pi}}
e^{\frac{-5i}{4}}
\left[
e^{\frac{5i}{2}}
\text{erfi}\left(\frac{\sqrt{i}\left[ 2\ln(x) - i \right]}{2} \right) 
 -  
\text{erf}\left(\frac{\sqrt{i}\left[ 2\ln(x) + i \right]}{2} \right)\right]
\right)
\end{align*}

A quick check in Mathematica gives: 
f[x_] := Log[x] Cos[1 + (Log[x])^2]
FullSimplify[Integrate[f[x], x]]

1/8 ((-1)^(1/4) E^(-((5 I)/4))
  Sqrt[\[Pi]] (-Erf[1/2 (-1)^(1/4) (I + 2 Log[x])] + 
  E^((5 I)/2) Erfi[1/2 (-1)^(1/4) (-I + 2 Log[x])]) + 
  4 x Sin[1 + Log[x]^2])

Which maybe should have been the whole answer...

See also this question, which solves a similar problem.
