evaluation of real $k$ for which Integral Converges

Question

Finding real value of $k$ for which the integral $$\int^{\infty}_{0}3^{(3+3k)x}dx$$ converges and finding its value, is

What i try

$\bullet$ If $3+3k=0$. Then $$I =\int^{\infty}_{0}1\cdot dx\rightarrow \infty(\text{Diverge})$$

$\bullet$ If $3+3k\neq 0$ Then Let $$I =\int^{\infty}_{0}3^{(3+3k)x}dx=\frac{3^{(3+3k)x}}{\ln(3+3k)}\bigg|^{\infty}_{0}$$

Now i will form Subcases

$\bullet$ For $3+3k>0$. We have $$I \rightarrow \infty(\text{Diverge})$$

$\bullet$ For $3+3k<0$. Then $\ln(3+3k)$ is Not defined.

Means no real values of $k$ for which Integral Converges.

Is my process is Right. If not How do i solve it. Help me please

Answer 1

$$\int^{\infty}_{0}3^{(3+3k)x}dx = \frac{3^{(3+3k)x}}{(3+3k)\ln 3}\bigg|^{\infty}_{0}$$

and not what you wrote. Apart from that, your approach is correct.

Answer 2

If $k=-1,$ the given integral diverges. Otherwise, $$\int^b_{0}3^{(3+3k)x}dx$$ $$=\int^b_{0}e^{\ln 3(3+3k)x}dx$$ $$=\frac{3^{(3+3k)x}}{\ln3(3+3k)}|_0^b$$ Thus the integral converges iff $k<-1.$