I need help with this integral, I'm not sure if this integral even has a solution.
$$\text{I}=\int_{-1}^{1}e^{-((\frac{1}{t-1})^2+(\frac{1}{t+1})^2)}\text{d}t$$
the graph of the integrand: Not able to post the picture, so here's a link instead
The approximate numeric value of the integral:
$$\text{I}\approx\text{0.0843896007479...}$$
I've tried to evaluate it, but I've yet to yield a solution.
An attempt:
$\text{I}=\int_{-1}^{1}e^{-((\frac{1}{t-1})^2+(\frac{1}{t+1})^2)}\text{d}t=2\int_{0}^{1}e^{-2\bigl(\frac{t^2+1}{(t-1)^2}\bigr)}\text{d}t$
First I substituted $t=\sqrt{1-x^2}\Rightarrow\text{d}t=\frac{-x}{\sqrt{1-x^2}}\text{d}x$ and got
$=2\int_{0}^{1}e^{-2\bigl(\frac{2-x^2}{x^4}\bigr))}\frac{x}{\sqrt{1-x^2}}\text{d}x=2\int_{0}^{1}e^{-\Bigr(\bigr(\frac{2}{x^2}\bigl)^2-\bigr(\frac{2}{x^2}\bigl)\Bigl)}\frac{x}{\sqrt{1-x^2}}\text{d}x$
Then I substituted $\frac{2}{x^2}=u\Rightarrow\text{d}x=-\sqrt{\frac{2}{u^3}}\text{d}u$ and got
$=2\int_{2}^{\infty}\frac{e^{-u^2+u}}{\sqrt{u^3}\sqrt{u-2}}\text{d}u=2\sqrt[4]{e}\int_{2}^{\infty}\frac{e^{-\bigl(u-\frac{1}{2}\bigr)^2}}{\sqrt{u^3}\sqrt{u-2}}\text{d}u=2\sqrt[4]{e}\int_{2}^{\infty}e^{-\bigl(u-\frac{1}{2}\bigr)^2}\text{d}\bigl(\sqrt{\frac{u-2}{u}}\bigr)$
Using integration by part, I obtain the following form(the first part goes to zero.)
$=4\sqrt[4]{e}\int_{2}^{\infty}e^{-\bigl(u-\frac{1}{2}\bigr)^2}\bigl(u-\frac{1}{2}\bigr)\sqrt{\frac{u-2}{u}}\text{d}u$
Last substitution before I hit obstacle, $u-1=v\Rightarrow\text{d}u=\text{d}v$
$=4\sqrt[4]{e}\int_{1}^{\infty}\sqrt{\frac{u-1}{u+1}}\bigl(u+\frac{1}{2}\bigr)e^{-\bigl(u+\frac{1}{2}\bigr)^2}\text{d}u$
Then I was stuck. I've tried to proceed after this, but it didn't go anywhere.
Worth mentioning:
There're several forms of this integral that I deemed are probably useful or convenient.
$$I=4\sqrt[4]{e}\int_{1}^{\infty}\sqrt{\frac{x-1}{x+1}}\bigl(x+\frac{1}{2}\bigr)e^{-\bigl(x+\frac{1}{2}\bigr)^2}\text{d}x$$ $$=4\sqrt[4]{e}\int_{\frac{3}{2}}^{\infty}\sqrt{\frac{2x-3}{2x+1}}xe^{-x^2}\text{d}x$$ $$=2\sqrt[4]{e}\int_{\frac{9}{4}}^{\infty}\sqrt{1-4\Bigr(\frac{2\sqrt{x}-1}{4x-1}\Bigl)}e^{-x}\text{d}x$$ $$=\frac{1}{e^2}\int_{0}^{\infty}\sqrt{\frac{(\sqrt{4x+9}-2)^2-1}{x+2}}e^{-x}\text{d}x$$
That's all I have done, Is this integral has a solution? Genuinely wondering.
This is not an answer.
I am more than skeptical about a possible closed form.
If you build the $[4,4]$ Padé approximant around $t=0$ $$e^{-\frac{1}{(t+1)^2}-\frac{1}{(t-1)^2}}=\frac 1{e^2}\times \frac{1-\frac{191 }{31}t^2+\frac{323 }{31}t^4 } {1-\frac{5 }{31}t^2+\frac{45 }{31}t^4 }$$ which is not too bad for $-\frac 12 \le x \le \frac 12$ and you integrate between these bounds, you have an ugly closed form expression which is $0.0834826$ while, for this range, numerical integration gives $0.0832537$.
The problem is now to approximate the tails and I am short in ideas.